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Showing two things are equal by Fourier series Given the Fourier series for the function:
$$f(x) = x+\frac14x^2 \quad -\pi\leq x \lt \pi$$
$$f(x)=f(x+2\pi) \quad -\infty \leq x \lt \infty$$
is
$$\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \left(\frac{\cos(nx)}{n^2}-\frac{2\sin(nx)}{n}\right)$$
show that$$\sum \limits_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$
Now I hope this doesn't mean I have to calculate the Fourier series for this, since it seems to be given, since I had written down the Fourier coefficient for $a_0$ and most of $a_n$ and realised that this will simply be too time consuming in an exam!
If I sub in $x=2\pi$ for $f(2\pi)=2\pi + \pi^2$ and the fourier gives:
$$\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \left(\frac{\cos(2n\pi)}{n^2}-\frac{2\sin(2n\pi)}{n}\right)=\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \left(\frac{1}{n^2}\right)$$
$$2\pi + \pi^2=\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \frac{1}{n^2}$$
$$2\pi+\frac{11\pi^2}{12}=\sum \limits_{n=1}^\infty (-1)^n \frac{1}{n^2}$$
So this isn't right. Perhaps another value for $x$, how about $x=\pi$
$$\pi + \frac{\pi^2}{4}=\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \left(\frac{\cos(n\pi)}{n^2}-\frac{2\sin(n\pi)}{n}\right)$$
$$\pi + \frac{\pi^2}{4}=\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \left(\frac{(-1)^n}{n^2}\right)$$
$$\pi+ \frac{\pi^2}{6}=\sum \limits_{n=1}^\infty\frac{1}{n^2}$$
Almost correct, but I have an extra $\pi$ term. Is there a way I could have gotten this on the first guess, and how do I show it?
| Try $x = \pi$. When $x=\pi$, by Dirichlet Theorem (will look up which one),
$$
f(\pi) = \frac{f(-\pi)+f(\pi)}{2} = \frac{\pi^2}{4}
$$
since you have a jump discontinuity at $x=\pi$ (see theorem).
This gives you
$$
\sum_{n=1}^{\infty}\frac{1}{n^2} + \frac{\pi^2}{12}=\frac{\pi^2}{4}
$$
Theorem: Let $f$ be periodic and piecewise differentiable. Then at each point $\theta$ the symmetric partial sum
$$
S_N(\theta) = \sum_{-N}^Na_ne^{in\theta}
$$
converges to $\frac{1}{2}(f(\theta +) + f(\theta -))$; if $f$ is continuous at $\theta$, then it converges to $f(\theta)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Factorizing Given Problem I have searched through various site's and forums but couldn't find the answer to my problem,
$$z^2-\frac{1}{2}z-\frac{1}{4}=0$$
How will you factorize this
As I can't find $2$ numbers that give me $ac = -\frac{1}{4}$ and $b = -\frac{1}{2}$ when added
Is there a different approach for such sums.``
| The quadratic formula can be used to find the roots of a quadratic equation.
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$ for a quadratic of the form $ax^2+bx+c=0$.
So for $a=1,b=-\frac{1}{2},c=-\frac{1}{4}$, we have that
$x=\dfrac{-\left(\frac{1}{2}\right)\pm\sqrt{(-\frac{1}{2})^2-4(1)(-\frac{1}{4})}}{2(1)}=\dfrac{-\frac{1}{2}\pm\sqrt{\frac{1}{4}+1}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I evaluate $\lim_{n \to\infty}\left(1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)\right)/\left(1^2+2^2+3^2+\dots+n^2\right)^2$? How can I evaluate this limit? Give me a hint, please.
$$\lim_{n \to\infty}\frac{1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)}{\left(1^2+2^2+3^2+\dots+n^2\right)^2}$$
| The numerator is bounded above by $n.n(n+1)(n+2)$, fourth degree. The denominator is exactly $(2n^3+3n^2+n)^2/36$ (from the square pyramidal number formula), sixth degree. So the denominator "wins".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Sum of $\sum_{n=0}^{\infty }\frac{1}{4^{(n/3)+1}}$ Find the sum of $$\sum_{n=0}^{\infty }\frac{1}{4^{(n/3)+1}}$$
| geometric sequence with $$q=\frac{1}{4^{\frac{1}{3}}}\\$$
$$\frac{1}{4^{1}}+\frac{1}{4^{\frac{4}{3}}}++\frac{1}{4^{\frac{7}{3}}}++\frac{1}{4^{\frac{10}{3}}}+ ...\\=\frac{a_{1}}{1-q}=\\=\frac{\frac{1}{4^{1}}}{1-\frac{1}{4^{\frac{1}{3}}}}=\\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I express the sum $(1+k)+(1+k)^2+\ldots+(1+k)^N$ for $|k|\ll1$ as a series? Wolfram Alpha provides the following exact solution
$$ \sum_{i=1}^N (1+k)^i = \frac{(1+k)\,((1+k)^N-1)}{k}.$$
I wish to solve for $N$ of the order of several thousand and $|k|$ very small (c. $10^{-12}).$ When I do this on a computer in excel the software cannot handle it (because of truncation of significant figures) and the results are nonsense.
I thought to approximate the result using the first few terms of a series in increasing powers of $k$. I can multiply out the first few terms and examine the patterns in the following pyramid...
$$(1+k)^1 = k +1 $$
$$(1+k)^2 = k^2 +2k +1 $$
$$(1+k)^3 = k^3 +3k^2 +3k +1 $$
$$(1+k)^4 =k^4+4 k^3+6 k^2+4 k+1$$
$$(1+k)^5 =k^5+5 k^4+10 k^3+10 k^2+5 k+1$$
$$(1+k)^6 =k^6+6 k^5+15 k^4+20 k^3+15 k^2+6 k+1$$
So for example, for $N=3$ we would obtain the sum
$$
S= k^3 +4k^2 +6k +1
$$
The results suggest a solution with a pattern of the form
$$
S = a + bk^1 +ck^2+dk^3...
$$
I can see that $a=N$. The other coefficients increase monotonously and it might be possible to determine a formula for the coefficients from the pattern. Although the general pattern is not convergent, it is possible that for certain restricted ranges of $N$ and $k$ a convergent formula could be obtained. If so then it is possible that a useful approximation of S can be obtained just by evaluating the first few terms in the series.
But is there a well-known general formula for the terms in this series or can one be derived algebraically from the original formula?
UPDATE
following on from the answer by User73985...
$$ S=\sum_{i=1}^N (1+k)^i = N + \sum_{j=2}^{N+1}\binom{N+1}{j}k^{j-1}$$
So
$$ S= N + \sum_{j=2}^{N+1}\frac{(N+1)!}{(N+1-j)!j!}k^{j-1}$$
then
$$ S= N
+ \frac{(N+1)!}{(N-1)!2!} k^{1}
+ \frac{(N+1)!}{(N-2)!3!} k^{2}
+ \frac{(N+1)!}{(N-3)!4!} k^{3} +...
$$
giving
$$ S= N
+ \frac{(N+1)(N)}{2!} k^{1}
+ \frac{(N+1)(N)(N-1)}{3!} k^{2}
+ \frac{(N+1)(N)(N-1)(N-2)}{4!} k^{3} +...
$$
thus
$$ S= N
+ \frac{N^2+N}{2} k^{1}
+ \frac{N^3-N}{6} k^{2}
+ \frac{N^4-2 N^3-N^2+2 N}{24} k^{3} +...
$$
For $N=1 to 10,000$ and $k= 2.40242 * 10^{-12}$ this formula can be truncated to
$$ S = N + \frac{N^2+N}{2} k^{1}
$$
and then gives results very close to those expected. Because $k$ is so small relative to $n$ the terms in higher powers of $k$ can be ignored. Note that the coefficient of $k^1$ is consistent with that found by examination of the coefficients in the "pyramid" presented above.
| An alternative approach using binomial coefficients:
$$\begin{align}
\sum_{i=1}^{N}(1+k)^i&=\sum_{i=1}^N\sum_{r=0}^i \binom ir k^r\\
&=k^0\sum_{i=1}^N\binom i0+\sum_{r=1}^Nk^r\sum_{i=r}^N\binom ir\\
&=N+\sum_{r=1}^Nk^r\binom{N+1}{r+1}
\end{align}$$
NB - the result is the same as
$$\qquad N+\sum_{r=2}^{N+1}k^{r-1}\binom {N+1}r$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $ \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+\cdots + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$ for $n\in \mathbb N$ I want to prove that if $n \in \mathbb N$ then $$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}.$$
I think I am stuck on two fronts. First, I don't know how to express the leading terms on the left hand side before the $\dfrac{n}{(n+1)!}$ (or if doing so is even necessary to solve the problem). I am also assuming that the right high side should initially be expressed $1 - \dfrac{1}{(n+2)!}$. But where to go from there.
I'm actually not sure if I'm even thinking about it the right way.
| Hint. If, for somespecific $n$, we have
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}\ ,$$
then
$$\eqalign{\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n+1}{(n+2)!}
&=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}\cr
&=1-\frac{n+2}{(n+2)!}+\frac{n+1}{(n+2)!}\cr
&=1-\frac{1}{(n+2)!}\ .\cr}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Double Integral $\int\limits_0^a\int\limits_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$ How to solve this integral?
$$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$
my attempt
$$
\int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}=
\int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\
\rho^2=y^2+a^2\\
x=\rho\tan\theta\\
dx=\rho\sec^2\theta \, d\theta\\
x^2+\rho^2=\rho^2\sec^2\theta\\
\int_0^a\!\!\!\int_0^{\arctan\frac{a}{\rho}}\frac{\rho\sec\theta}{\rho^3\sec^3\theta}d\theta \, dy=
\int_0^a\!\!\!\frac{1}{\rho^2}\!\!\!\int_0^{\arctan\frac{a}{\rho}}\cos\theta \, d\theta \, dy=\\
\int_0^a\frac{1}{\rho^2}\sin\theta\bigg|_0^{\arctan\frac{a}{\rho}} d\theta \, dy=
\int_0^a\frac{1}{\rho^2}\frac{x}{\sqrt{x^2+\rho^2}}\bigg|_0^ady=\\
\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy$$
Update:
$$\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy=\frac{\pi}{6a}$$
| Let $y=\sqrt{a^2+x^2} \tan \theta$, $dy=\sqrt{a^2+x^2}~ \mbox{sec}^2 \theta~ d\theta$ Then
$$\int_{0}^{a} \int_{0}^{a} \frac{dx ~ dy} {(a^2+x^2+y^2)^{3/2}} =\int_{0}^{a}\frac{dx}{a^2+x^2}\int_{0}^{\tan^{-1}(a/\sqrt{a^2+x^2})} \cos \theta ~d\theta$$ $$ = a \int_{0}^{a} \frac{dx}{(x^2+a^2)\sqrt{(2a^2+x^2)}}.$$
$$ \Rightarrow I=\int_{0}^{a}\frac{1}{2i} \frac{dx}{\sqrt{2a^2+x^2}} \left [ \frac{1}{x-ia}-\frac{1}{x+ia}\right]=\Im \left( \int_{0}^{a}\frac{dx}{(x-ia)\sqrt{2a^2+x^2}}\right).$$
Use $(x-ia)=1/t$, then
$$I=-\Im \left (\int_{i/a}^{(1+i)/(2a)} \frac{dt}{\sqrt{a^2t^2+2iat+1}}\right) =-\Im \int_{2i}^{(1+3i)/2} \left(\frac{dv}{a\sqrt{v^2+(\sqrt{2})^2}}\right).$$
Letting $v=\sqrt{2} \tan \phi$, we find that $$I=-\Im \left( a^{-1} \ln \left[ \frac{1+3i+\sqrt{6i}}{2i \sqrt{2}(1+\sqrt{2})}\right]\right)= \frac{\pi}{6a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Integrate $\frac{\lambda y^{2}}{\sqrt{2\pi}} e^{-(\frac{1}{2}+ \lambda x)y^{2}}$ with respect to $y$ $$F(x,y)= \frac{\lambda y^{2}}{\sqrt{2\pi}} e^{-(\frac{1}{2}+ \lambda x)y^{2}}$$
Please show that the function when integrated with respect to $y$ is
$F_X(x)= \frac{\lambda}{(2\sqrt{2}(\frac{1}{2}+ \lambda x)^\frac{3}{2})}$ using $u =(\frac{1}{2}+ \lambda x)y^{2}$ and the gamma function.
| So, the two main passages are the change of variable and the gamma function. I'm supposing you want to integrate it between $0$ and $+\infty$.Therefore:
\begin{equation}
F_x(x) = \int_0^{+\infty} \frac{\lambda y^2}{\sqrt{2\pi}} e^{-(\frac{1}{2} + \lambda x)y^2} dy= \int_0^{+\infty} \frac{\lambda}{\sqrt{2\pi}} \frac{1}{(\frac{1}{2} + \lambda x)} (\frac{1}{2} + \lambda x) y^2 e^{-(\frac{1}{2} + \lambda x)y^2} dy =
\end{equation}
Let's change the variable by imposing $u=(\frac{1}{2} + \lambda x)y^2$ so that we have:
\begin{align}
& du= 2(\frac{1}{2} + \lambda x)y dy\\[5pt]
& y = \sqrt{\frac{u}{(\frac{1}{2} + \lambda x)}}
\end{align}
Applying the variable change leads to:
\begin{align}
Fx(x) &= \frac{\lambda}{\sqrt{2\pi}} \frac{1}{(\frac{1}{2} + \lambda x)} \int_0^{+\infty} u\,\, e^{-u} \frac{1}{2(\frac{1}{2} + \lambda x)\,y(u)} du \\[10pt]
&= \frac{\lambda}{\sqrt{2\pi}} \frac{1}{(\frac{1}{2} + \lambda x)} \int_0^{+\infty} u\,\, e^{-u}\frac{\sqrt{\frac{1}{2} + \lambda x}}{2(\frac{1}{2} + \lambda x)\,\sqrt{u}} du \\[5pt]
&= \frac{\lambda}{2\sqrt{2\pi} (\frac{1}{2} + \lambda x)^{3/2}} \int_0^{+\infty} u^{\frac{1}{2}}\,\, e^{-u} \,\,du
\end{align}
The latter integral is the gamma function of $\frac{3}{2}$ as the gamma function is defined as:
\begin{equation}
\Gamma(z) = \int_0^{+\infty} t^{z-1} e^{-t} dt
\end{equation}
Since that $\Gamma(\frac{3}{2})= \frac{\pi}{2}$ (the reason is not easy to prove, but you can retrieve those particular values from the wikipedia page) you have that:
\begin{equation}
Fx(x) = \frac{\lambda}{2\sqrt{2\pi} (\frac{1}{2} + \lambda x)^{3/2}} \frac{\sqrt{\pi}}{2} = \frac{\lambda}{4\sqrt{2} (\frac{1}{2} + \lambda x)^{3/2}}
\end{equation}
I still have a $\frac{1}{2}$ difference with your solution and honestly at the moment I can't figure out exactly where the mistake is, but I hope it can be still a useful reply.
| {
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"url": "https://math.stackexchange.com/questions/1033820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Characteristics method applied to the PDE $u_x^2 + u_y^2=u$ I am trying to solve: $u_x^2 + u_y^2=u$ with boundary conditions: $u(x,0)=x^2$. Unfortunately it leads to equations that makes no sense (sum of squares is $0$ and all constants are $0$). I would be grateful for a reasonable explanation how to solve that with help of characteristics.
| Let $u=v^2$ ,
Then $u_x=2vv_x$
$u_y=2vv_y$
$\therefore(2vv_x)^2+(2vv_y)^2=v^2$ with $v(x,0)=x$
$4v^2(v_x)^2+4v^2(v_y)^2=v^2$ with $v(x,0)=x$
$v_x^2+v_y^2=\dfrac{1}{4}$ with $v(x,0)=x$
$v_y^2=\dfrac{1}{4}-v_x^2$ with $v(x,0)=x$
$v_y=\pm\sqrt{\dfrac{1}{4}-v_x^2}$ with $v(x,0)=x$
$v_{xy}=\mp\dfrac{v_xv_{xx}}{\sqrt{\dfrac{1}{4}-v_x^2}}$ with $v(x,0)=x$
Let $w=v_x$ ,
Then $w_y=\mp\dfrac{ww_x}{\sqrt{\dfrac{1}{4}-w^2}}$ with $w(x,0)=1$
$w_y\pm\dfrac{ww_x}{\sqrt{\dfrac{1}{4}-w^2}}=0$ with $w(x,0)=1$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$
$\dfrac{dw}{dt}=0$ , letting $w(0)=w_0$ , we have $w=w_0$
$\dfrac{dx}{dt}=\pm\dfrac{w}{\sqrt{\dfrac{1}{4}-w^2}}=\pm\dfrac{w_0}{\sqrt{\dfrac{1}{4}-w_0^2}}$ , letting $x(0)=f(w_0)$ , we have $x=\pm\dfrac{w_0t}{\sqrt{\dfrac{1}{4}-w_0^2}}+f(w_0)=\pm\dfrac{wy}{\sqrt{\dfrac{1}{4}-w^2}}+f(w)$ , i.e. $w=F\left(x\mp\dfrac{wy}{\sqrt{\dfrac{1}{4}-w^2}}\right)$
$w(x,0)=1$ :
$F(x)=1$
$\therefore w=1$
$v_x=1$
$v(x,y)=x+g(y)$
$v_y=g_y(y)$
$\therefore1^2+(g_y(y))^2=\dfrac{1}{4}$
$(g_y(y))^2=-\dfrac{3}{4}$
$g_y(y)=\pm\dfrac{i\sqrt3}{2}$
$g_y=\pm\dfrac{i\sqrt3y}{2}+C$
$\therefore v(x,y)=x\pm\dfrac{i\sqrt3y}{2}+C$
$v(x,0)=x$ :
$C=0$
$\therefore v(x,y)=x\pm\dfrac{i\sqrt3y}{2}$
Hence $u(x,y)=\left(x\pm\dfrac{i\sqrt3y}{2}\right)^2=x^2\pm i\sqrt3xy-\dfrac{3y^2}{4}$
| {
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"url": "https://math.stackexchange.com/questions/1033906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $(b-a)\cdot f(\frac{a+b}{2})\le \int_{a}^{b}f(x)dx$ Let $f$ be continuously differentiable on $[a,b]$. If $f$ is concave up, prove that
$$(b-a)\cdot f\left(\frac{a+b}{2}\right)\le \int_{a}^{b}f(x)dx.$$
I know that (and have proved) $$(b-a)\cdot f\left(\frac{a+b}{2}\right)= \int_{a}^{b}f(x)dx$$ for any linear function on $[a,b]$. Also, the graph of $f$ lies above the tangent line at $(\frac{a+b}{2},f(\frac{a+b}{2}))$.
| Let $h(x)=f(x)-m x$ where $m=f'(\frac{b+a}2)$. By the Mean Value Theorem for integrals (also related to average value of a function), there is $c\in(a,b)$ such that $h(c)(b-a)=\int_{a}^{b}h(x)dx$. Note that $h$ is concave up and $h'(\frac{b+a}2)=0$ hence $h$ has a global (over $[a,b]$) minimum at $\frac{b+a}2$. Hence $h(c)\ge h(\frac{b+a}2)$. In other words $\frac1{b-a}\int_{a}^{b}h(x)dx\ge h(\frac{b+a}2)$. Thus $\frac{1}{b-a}\int_{a}^{b}h(x)dx = \frac1{b-a}(\int_{a}^{b}f(x)dx - \frac{m}{2} (b^2-a^2))= \frac1{b-a}\int_{a}^{b}f(x)dx - {m}\frac{b+a}{2}\ge$ $h(\frac{b+a}2)=f(\frac{b+a}2)-{m}\frac{b+a}{2}$
from which we obtain $\frac1{b-a}\int_{a}^{b}f(x)dx\ge f(\frac{b+a}2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$. Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$.
My solutions: the homogeneous portion is $a_n=c3^n$, and the inhomogeneous portion is $a^*_n=-1/2n^2-3/4n+9/8$.
This results in a final recurrence relation of
$$a_n=-\frac{1}{8}3^n-\frac{1}{2}n^2-\frac{3}{4}n+\frac{9}{8}.$$
I am just wondering if someone could check my work to make sure I have the procedure correct.
| Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the shifted recurrence by $ẓ^n$ and sum over $n \ge 0$, recognize some sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 1} z^n
&= 3 \sum_{n \ge 0} a_n z^n + \sum_{n \ge 0} (n + 1)^2 - 3 \sum_{n \ge 0} z^n \\
\frac{A(z) -a_0}{z}
&= 3 A(z) + \sum_{n \ge 0} n^2 z^n + 2 \sum_ {n \ge 0} n z^n - 2 \sum_{n \ge 0} z^n
\end{align*}$
You know the geometric sum:
$\begin{align*}
\sum_{n \ge 0} z^n
&= \frac{1}{1 - z}
\frac{d}{d z} \sum_{n \ge 0} z^n
&= \sum_{n \ge 0} n z^{n - 1} \\
z \frac{d}{d z} \sum_{n \ge 0} z^n
&= \sum_{n \ge 0} n z^n \\
&= \frac{z}{(1 - z)^2}
\sum_{n \ge 0} n^2 z^n
&= \frac{z + z^2}{(1 - z)^3}
\end{align*}$
Plug this in above, solve for $A(z)$, and write as partial fractions:
$\begin{align*}
A(z)
&= \frac{1 - z - z^2 - z^3}{(1 - z) (1 + z) (1 - 3 z) (1 + z + z^2)} \\
&= \frac{7 + 2 z}{39 (1 + z + z^2)}
+ \frac{21}{52 (1 - 3 z)}
+ \frac{1}{4 (1 + z)}
+ \frac{1}{6 (1 - z)} \\
&= \frac{(7 + 2 z) (1 - z)}{39 (1 - z^3)}
+ \frac{21}{52 (1 - 3 z)}
+ \frac{1}{4 (1 + z)}
+ \frac{1}{6 (1 - z)} \\
&= \frac{7 - 5 z - 2 z^2}{39 (1 - z^3)}
+ \frac{21}{52 (1 - 3 z)}
+ \frac{1}{4 (1 + z)}
+ \frac{1}{6 (1 - z)}
\end{align*}$
You want the coefficient of $z^n$ from this. Everything in sight is just a geometric series:
$\begin{align*}
[z^n] \left(
\frac{7 - 5 z - 2 z^2}{39 (1 - z^3)}
+ \frac{21}{52 (1 - 3 z)}
+ \frac{1}{4 (1 + z)}
+ \frac{1}{6 (1 - z)}
\right)
&= (7 [z^n] - 5 [z^{n - 1}] - 2 [z^{n - 2}]) \frac{1}{1 - z^3)}
+ \frac{21}{52} \cdot 3^n
+ \frac{1}{4} \cdot (-1)^n
+ \frac{1}{6} \\
&= [n = 0] \cdot 7
+ \frac{21}{52} \cdot 3^n
+ \frac{1}{4} \cdot (-1)^n
+ \frac{1}{6}
\end{align*}$
Here $[n = 0]$ uses Iverson's convention: $1$ if the condition is true, $0$ if false. If $n = 0$, the only relevant term is the first one, if $n > 0$ the other two terms cancel with it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Sinha's Theorem for Equal Sums of Like Powers $x_1^7+x_2^7+x_3^7+\dots$ Sinha’s theorem can be stated as, excluding the trivial case $c = 0$, if,
$$(a+3c)^k + (b+3c)^k + (a+b-2c)^k = (c+d)^k + (c+e)^k + (-2c+d+e)^k\tag{1} $$
for $\color{blue}{\text{both}}$ $k = 2,4$ then,
$$a^k + b^k + (a+2c)^k + (b+2c)^k + (-c+d+e)^k = \\(a+b-c)^k + (a+b+c)^k + d^k + e^k + (3c)^k
\tag{2}$$
for $k = 1,3,5,7$.
The system $(1)$ be equivalently expressed as,
$$\begin{align}
x_1^k+x_2^k+x_3^k\, &= y_1^k+y_2^k+y_3^k,\quad \color{blue}{\text{both}}\; k = 2,4\\
x_1+x_2-x_3\, &= 2(y_1+y_2-y_3)\\
x_1+x_2-x_3\, &\ne 0\tag{3}
\end{align}$$
There are only two quadratic parameterizations known so far to $(3)$, namely,
$$(-5x+2y+z)^k + (-5x+2y-z)^k + (6x-4y)^k = \\(9x-y)^k + (-x+3y)^k + (16x-2y)^k\tag{4}$$
where $126x^2-5y^2 = z^2$ and,
$$(6x+3y)^k + (4x+9y)^k + (2x-12y)^k = \\(-x+3y+3z)^k + (-x+3y-3z)^k + (-6x-6y)^k\tag{5}$$
where $x^2+10y^2 = z^2$ found by Sinha and (yours truly). The square-free discriminants are $D = 70, -10$, respectively.
Question: Any other solution for $(3)$ in terms of quadratic forms?
P.S. There are a whole bunch of elliptic curves that can solve $(3)$.
| I think not to introduce additional equations, and directly solve the system of equations.
$$\left\{\begin{aligned}&R^2+Q^2+T^2=X^2+Y^2+Z^2\\&R^4+Q^4+T^4=X^4+Y^4+Z^4\end{aligned}\right.$$
Using integer parameters $k,s,t$ - Will make a replacement.
$$a=3(k+s-t)^2+k(k-t)$$
$$b=3(k+s-t)^2+s(s-t)$$
$$c=3(k+s-t)^2-t^2+(k+s)t-2ks$$
$$x=3(k+s-t)^2-ks$$
$$y=3(k+s-t)^2-t^2+(k+s)t-ks$$
$$z=3(k+s-t)^2+k^2+s^2-(k+s)t$$
Then the solution can be written as:
$$R=3a^4+(4a-b)b^3+(4a-c)c^3-(4a-x)x^3-(4a-y)y^3-(4a-z)z^3$$
$$Q=(4b-a)a^3+3b^4+(4b-c)c^3-(4b-x)x^3-(4b-y)y^3-(4b-z)z^3$$
$$T=(4c-a)a^3+(4c-b)b^3+3c^4-(4c-x)x^3-(4c-y)y^3-(4c-z)z^3$$
$$X=(4x-a)a^3+(4x-b)b^3+(4x-c)c^3-3x^4-(4x-y)y^3-(4x-z)z^3$$
$$Y=(4y-a)a^3+(4y-b)b^3+(4y-c)c^3-(4y-x)x^3-3y^4-(4y-z)z^3$$
$$Z=(4z-a)a^3+(4z-b)b^3+(4z-c)c^3-(4z-x)x^3-(4z-y)y^3-3z^4$$
To obtain relatively Prime solutions - after substitution should be reduced to common divisor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Irrational number inequality : $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$ it is easy and simple I know but still do not know how to show it (obviously without simply calculating the sum but manipulation on numbers is key here.
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$$
| Or, one can consider the sum $\sqrt{3} + \sqrt{3/2} + 1$; we have
$\sqrt{3} > 1, \tag{1}$
$\sqrt{\dfrac{3}{2}} > 1, \tag{2}$
$1 \ge 1, \tag{3}$
(this last inequality being particularly subtle!) and adding (1)-(3) we have
$\sqrt{3} + \sqrt{\dfrac{3}{2}} + 1 > 3; \tag{4}$
now we divide by $\sqrt{3}$, and voila!, we obtain
$1 + \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{3}} > \sqrt{3}, \tag{5}$
the long-sought result!
With a tip-o'-the-hat to Eric Stucky, tho' I saw not his answer 'til mine nearly written was! (Yodaspeak!)
Note Added 24 November 2014, 1:02 PM PST: We can generalize: since
$\sqrt{\dfrac{n}{k}} > 1, \; \; 1 \le k < n, \tag{6}$
we can, again invoking (3), write
$\sum_{k = 1}^n \sqrt{\dfrac{n}{k}} > n, \tag{7}$
and now dividing by $\sqrt{n}$ yields
$\sum_1^n \dfrac{1}{\sqrt{k}} > \sqrt{n}. \tag{8}$
End of Note.
Hope this helps. Cheers,
and as ever,
Fiat Lux!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
Find equation of Tangent line at $(4, 1)$ on $5y^3 + x^2 = y + 5x$ Can someone help me find equation of tangent line at $(4, 1)$ on
$5y^3 + x^2 = y + 5x$
$Y=f(x)$
I dont know how to isolate the $Y$
| knowing
$$
y-y_0 = y'(x-x_0)
$$
$$
\frac{d}{dx}[3y^3+x^2] = \frac{d}{dx}[y+5x]
$$
$$
15y^2 y' + 2x = y' + 5
$$
$$
2x-5 = y'[1-15y^2] \therefore
$$
$$
y' = \frac{2x-5}{1-15y^2}
$$
When at the point (4,1) you get
$$
y-1 = \frac{2(4)-5}{1-15}(x-4)
$$
$$
y = \frac{12-3x}{14}+1 = \frac{13}{7}-\frac{3x}{14}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving of $\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{144n^2-144n+35}$ This is a homework for my son, he needs the proving.I tried to solve it by residue theory but I couldn't.
$$\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{144n^2-144n+35}$$
| I wouldn't say it's very illuminating, but here's an answer:
We have
\begin{align*}
\sum_{n=1}^\infty \frac{14}{576n^2 - 576 + 95}
&= \sum_{n=1}^\infty \frac{1}{24n - 19} - \frac{1}{24n - 5} \\
&= \frac{1}{24} \sum_{n=1}^\infty \frac{1}{n - 19/24} - \frac{1}{n - 5/24} \\
&= \frac{1}{24} \left(\psi\left(\frac{19}{24}\right) - \psi\left(\frac{5}{24}\right)\right),
\end{align*}
where $\psi$ is the digamma function. (The last equality follows more or less directly from its definition, or at least one of its definitions.) Since $\psi$ happens to satisfy $\psi(1 - z) - \psi(z) = \pi \cot \pi z$ (cf. the corresponding formula for $\Gamma$), we therefore have
\begin{align*}
\sum_{n=1}^\infty \frac{14}{576n^2 - 576 + 95}
&= \frac{\pi}{24} \cot \frac{5\pi}{24}
\end{align*}
The same method gives
\begin{align*}
\sum_{n=1}^\infty \frac{1}{144n^2 - 144n + 35}
&= \frac{1}{2}\sum_{n=1}^\infty \frac{1}{12n - 7} - \frac{1}{12n - 5} \\
&= \frac{1}{24} \left(\psi\left(\frac{7}{12}\right) - \psi\left(\frac{5}{12}\right)\right) \\
&= \frac{\pi}{24}\cot \frac{5\pi}{12}.
\end{align*}
Thus the given sum is equal to
\begin{align*}
\frac{\pi}{24}\left(\cot \frac{5\pi}{24} - \cot \frac{5\pi}{12}\right) = \frac{\pi}{24}\left(\sqrt{6} - \sqrt{2}\right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
If $(1+x+x^2)^{25} =\sum^{50}_{r=0} a_r x^r$ then .... If $$(1+x+x^2)^{25} =\sum^{50}_{r=0} a_r x^r$$
then find :
$\sum^{16}_{r=0} a_{3r} =$
My approach :
let (1+x) =t therefore,
$(1+x+x^2)^{25} =\sum^{50}_{r=0} a_r x^r$
=$(1+x+x^2)^{25} = (t+x^2)^{25}$
$= ^{25}C_0t^{25} +^{25}C_1 t^{24}x^2 +^{25}C_2 t^{23}x^4 +\cdots + x^{50}$
Also expanding the right hand side which is $\sum^{50}_{r=0} a_r x^r = a_0x^0 +a_1x+a_2x^2 +\cdots a_{50}x^{50}$
But unable to correlate this with the given problem i.e how to find $\sum^{16}_{r=0} a_{3r}$ please guide further on this. thanks.
| Hint: Evaluate at the two roots of the equation $1+x+x^2=0$ and at $x=1$, that is, at the three cube roots of unity, and add. The key fact is that if $\omega$ is one of the roots of $1+x+x^2=0$, then $\omega^2$ is the other, and $1^k+\omega^k+\omega^{2k}=0$ unless $k$ is divisible by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find a closed form for the equations $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$ This is the assignment I have:
Find a closed form for the equations
$1^3 = 1$
$2^3 = 3+5$
$3^3 = 7+9+11$
$4^3 = 13+15+17+19$
$5^3 = 21+23+25+27+29$
$...$
Hints. The equations are of the form $n^3 = a1 +a2 +···+an$, where
$a_{i+1} = a_i +2$ and $a_0 =n(n−1)+1$.
My reasoning:
We have to find a formula that give us $n^3$ summing operands. (why is this useful?)
We know that the first operand (or term) of the sum is $a_0 =n(n−1)+1$.
In fact, if you put $n = 3$, then $a_0 = 3(3 − 1) + 1 = 3*2 + 1 = 7$, which is exactly the first number of sum.
Then I notice that each $n$ sum has $n$ operands, and each operand differs from one another of 2.
Thus I came out with this formula:
$$
\sum\limits_{i=0}^{n-1} a_0 + 2 \cdot i
$$
where $a_0 =n(n−1)+1$
For example, if $n = 3$, then we have
$(n(n−1)+1 + 2 \cdot 0) + (n(n−1)+1 + 2 \cdot 1) + (n(n−1)+1 + 2 \cdot 2) \equiv$
$\equiv (7 + 0) + (7 + 2) + (7 + 4) \equiv$
$\equiv 7 + 9 + 11$
Which is what is written as third example.
I don't know if this is correct form or even if this is a closed form, that's why I am asking...
| $$\begin{align}
1^3&=1\\
2^3&=3+5\\
3^3&=7+9+11\\
4^3&=13+15+17+19\\
5^3&=21+23+25+27+29\\
\vdots &= \vdots\\
n^3&=[(n^2-n+1)]+[(n^2-n+1)+2]+[(n^2-n+1)+4]+\cdots+[(n^2-n+1)+2(n-1)]\\
&=\sum_{r=1}^n(n^2-n+1)+2(r-1)\qquad \blacksquare
\end{align}$$
That appears to be the formula required for the "series".
The formula expresses the cube of an integer ($n$) as the sum of $n$ integers which are in arithmetic progression with common difference of $2$ (and is not so much about the summation of the first $n$ cubes).
To show that this is correct:
$$\begin{align}
\text{RHS}&=\sum_{r=1}^n[\color{blue}{(n^2-n+1)}+2(r\color{blue}{-1})]\\
&=n(n^2-n-1)+2\sum_{r=1}^n r\\
&=n(n^2-n-1)+2\cdot \frac {n(n+1)}2\\
&=n^3=\text{LHS}\qquad \blacksquare
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluation of $\int \frac{x^4}{(x-1)(x^2+1)}dx$ Evaluation of $\displaystyle \int \frac{x^4}{(x-1)(x^2+1)}dx$
$\bf{My\; Try::}$ Let $$\displaystyle I = \int\frac{x^4}{(x-1)(x^2+1)}dx = \int \frac{(x^4-1)+1}{(x-1)(x^2+1)}dx = \int\frac{(x-1)\cdot (x+1)\cdot (x^2+1)}{(x-1)(x^2+1)}+\int\frac{1}{(x-1)(x^2+1)}dx$$
So $\displaystyle I = \int (x+1)dx+J\;\,\;,$ Where $\displaystyle J = \int\frac{1}{(x-1)(x^2+1)}dx$
Now can we solve $J$ without using Partial fraction.
If yes then plz explain me, Thanks
| Since we aren't permitted to use Partial Fraction, therefore undergo the following method,
$x=\tan \theta \implies dx=\sec^2 \theta\ d\theta$
$\therefore\displaystyle\int \dfrac{dx}{(x-1)\left(x^2+1\right)}=\displaystyle\int \dfrac{\sec^2 \theta\ d\theta}{(\tan \theta-1)\sec^2 \theta}=\displaystyle\int\dfrac{d\theta}{1-\tan \theta }$
$\hat{J}=\displaystyle\int\dfrac{d\theta}{1-\tan \theta },\ \hat{I}=\displaystyle\int\dfrac{\tan \theta}{1-\tan \theta }d\theta$
$$\hat{J}=\displaystyle\int\dfrac{1+\tan \theta}{1-\tan \theta }d\theta-\hat{I}\tag{1} $$
$$\hat{J}=\displaystyle\int d\theta+\hat{I}\tag{2} $$
$$\color{blue}{\therefore \hat{I}=\dfrac{1}{2}\left(\displaystyle\int\dfrac{1+\tan \theta}{1-\tan \theta }d\theta-\displaystyle\int d\theta\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$
Prove that $$\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$$
I tried to to prove the above statement using the AM-HM inequality:
$$\begin{align}\frac{1}{2^n - 2^{n-1}}\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} &\ge \frac{2^n - 2^{n-1}}{\sum_{i = 2^{n-1} + 1}^{2^n}(a + ib)}\\
\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} &\ge \frac{(2^n - 2^{n-1})^2}{\frac{2^n -2^{n-1}}{2}(2a + (2^n + 2^{n-1} + 1)b)}\\
&=\frac{2^{n+1} - 2^n}{2a + (2^n + 2^{n-1} + 1)b}\end{align}$$
after which I am more or less stuck. How can I continue on from here, or is there another method?
| For $a,b,x>0$, the function $f(x)=\frac{1}{a+bx}$ is convex. So by Jensen's inequality
$$
\frac{1}{2^{n-1}}\sum_{i=2^{n-1}+1}^{2^n}f(i)\geq f\left(\frac{1}{2^{n-1}}\sum_{i=2^{n-1}+1}^{2^n}i\right)=f(0.5+3\times2^{n-2})=\frac{1}{a+(0.5+3\times2^{n-2})b}.
$$
It remains to show $2^{n-1}$ times the rightmost expression above is greater than or equal to the RHS of the desired inequality. This amounts to computing:
$$
2^{n-1}(a+2b)-(a+(0.5+3\times2^{n-2})b)=a(2^{n-1}-1)+b(2^n-0.5-3\times2^{n-2})\\
=a(2^{n-1}-1)+b(2^{n-2}-0.5)=(2^{n-1}-1)(a+b/2)\geq 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving formula for sum of squares with binomial coefficient $$\sum_{k=0}^{n-1}(k^2)= \binom{n}{3} + \binom{n+1}{3}$$
How should I prove that it is the correct formula for sum of squares?
Should I use induction to prove the basis? Any help is appreciated.
| It's easy to prove this directly from the definition of the binomial coefficient. The advantage is you don't need to remember any binomial identities, or try to rewrite a fraction to look like a binomial.
The left-hand side is $$\sum_{k=0}^{n-1}k^2=\frac{n(n-1)(2n - 1)}{6}$$ by a well known formula (you can also prove this by induction very easily).
The right-hand side is $$\binom{n}{3} + \binom{n + 1}{3}=\frac{n!}{3!(n - 3)!} + \frac{(n + 1)!}{3!(n + 1 - 3)!} = \frac{n! \left(n - 2\right)! + \left(n - 3\right)! \left(n + 1\right)!}{6 \left(n - 3\right)! \left(n - 2\right)!}$$
Writing $n! = n(n - 1)(n - 2)(n - 3)!$ we can cancel $(n - 3)!$ giving $$\frac{n(n - 1)(n - 2) \left(n - 2\right)! + \left(n + 1\right)!}{6 \left(n - 2\right)!}.$$ Similarly we can write $(n + 1)! = (n + 1)n(n - 1)(n - 2)!$ and cancel $(n -2)!$ giving $$\frac{n(n - 1)(n - 2) + \left(n + 1\right)n(n - 1)}{6}.$$ Factoring $n(n - 1)$ in the numerator gives $$\frac{n(n - 1)(n - 2 + n + 1)}{6} = \frac{n(n - 1)(2n - 1)}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $ \int_{0}^{\infty}\frac{\ln (1+16x^2)}{1+25x^2}\mathrm d x$ how to solve such type of definite integration?
I would like to see various methods to evaluate following integral
$$
\int_{0}^{\infty}\frac{\ln (1+16x^2)}{1+25x^2}\mathrm d x$$
| Rewrite the integral as
$$ 4 \log{2}\int_0^{\infty} dx \frac{1}{1+25 x^2} + \int_0^{\infty} dx \frac{\log {\left(\frac1{16}+x^2 \right)}}{1+25 x^2} $$
Consider
$$I(a) = \int_0^{\infty} dx \frac{\log{(a+x^2)}}{1+25 x^2} $$
$$\begin{align}I'(a) &= \int_0^{\infty} dx \frac{1}{(a+x^2)(1+25 x^2)}\\ &= \frac12 \int_{-\infty}^{\infty} dx \frac{1}{(a+x^2)(1+25 x^2)} \\ &= \frac1{4 \pi} \int_{-\infty}^{\infty}dk \, \frac{\pi}{\sqrt{a}} \, e^{-|k|\sqrt{a}} \frac{\pi}{5} e^{-|k|/5}\\ &= \frac{\pi}{10 \sqrt{a}} \int_0^{\infty} dk \, e^{-\left (\sqrt{a}+\frac15 \right ) k}\\ &= \frac{\pi}{10 \sqrt{a}} \frac1{\sqrt{a}+\frac15}\end{align}$$
Thus
$$I(a)= \frac{\pi}{2} \int \frac{da}{5 a + \sqrt{a} } = \pi \int \frac{dy}{5 y+1} = \frac{\pi}{5} \log{\left (\sqrt{a}+\frac15 \right )} + C$$
We find the constant of integration by evaluating
$$I(0) = 2 \int_0^{\infty} dx \frac{\log{x}}{1+25 x^2} = -\frac{\pi}{5} \log{5} \implies C=0$$
The integral is therefore
$$\frac{2 \pi \log{2}}{5} + \frac{\pi}{5} \log{\frac{9}{20}} = \frac{\pi}{5} \log{\frac{9}{5}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Solve the linear homogeneous recurrence relation with constant coefficients $$9a_{n} = 6a_{n-1}-a_{n-2}, a_{0}=6, a_{1}=5$$
So
$$x^n = (6x^{n-1}-x^{n-2})\div9$$
thus
$$[x^2 = (6x-1)\div9] \equiv [x^2 - \frac{2}{3}x + \frac{1}{9} = 0], x=\frac{1}{3}$$
also
$$a_{2}=\frac{8}{3}, a_{3}=\frac{31}{27}$$
How do I plug in that x/root to solve for the given recurrence relation?
I tried
$$9a_{n} = 6(\frac{1}{3})^{n-1} - (\frac{1}{3})^{n-2}$$
which for n=1 gives 5... which is correct, for a_1 though, not 9a_1? So that's not right.
| Here are the steps
$$
a_{0}=6
$$
$$
a_{1}=5
$$
$$
9a_{n} = 6a_{n-1}-a_{n-2}
$$
Lets rewrite this recurrence as
$$
a_{n} = \frac69a_{n-1}-\frac19a_{n-2}
$$
$$
a_{n} = \frac23a_{n-1}-\frac1{3^2}a_{n-2}
$$
Now let's multiply by $3^n$
$$
3^0a_{0}=3^06=6
$$
$$
3^1a_{1}=3^15=15
$$
$$
3^na_{n} = 2\cdot 3^{n-1}a_{n-1}-3^{n-2}a_{n-2}
$$
Let $S_n=3^na_n$, then
$$
S_0=6
$$
$$
S_{1}=15
$$
$$
S_n= 2S_{n-1}-S_{n-2}
$$
It follows that
$$
S_n= S_1n-S_0(n-1)=9n+6=3(3n+2)
$$
Therefore
$$
a_n=\frac{3(3n+2)}{3^n}=3^{1-n}(3n+2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it would be shown that:
$\dfrac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$ is equivalent to $\dfrac{(k + 1)[3(k+1)+1]}2$
Any assistance would be appreciated.
| The formula must be a quadratic polynomial (because its first order difference is a linear polynomial) and has three independent coefficients. It suffices to identify for three different values of $n$:
$$\begin{align}1=\frac{1(3\cdot1-1)}2
\\1+4=\frac{2(3\cdot2-1)}2
\\1+4+7=\frac{3(3\cdot3-1)}2\end{align}$$
This completes the proof for any $n$ !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 1
} |
Show that $\prod_{i=2}^n \left(1-\frac{1}{i^2}\right) = \frac{n+1}{2n}$for $n \in \Bbb{N}$, $ n \ge 2$ Use mathematical induction to shoe that fpr any $n\in N$, if $n\ge2$, then
$$\prod_{i=2}^{n}\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2n}$$
So I understand what's happening up until the first red circle. Why does 1 - become $(n+1)^2$?
When expanding $(n+1)^2$ I thought it'd be $n^2+2$, why is there an extra $n+1$?
| $$\color{blue}{1}-\frac{1}{(n+1)^2}=\color{blue}{\frac{(n+1)^2}{(n+1)^2}}-\frac{1}{(n+1)^2}=\frac{(n+1)^2-1}{(n+1)^2}$$
Now use $$(a+b)^2=(a+b)(a+b)=a^2+ab+ab+b^2=a^2+2ab+b^2$$
to get $(n+1)^2=n^2+2n+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1051587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $ \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ Evaluation of $\displaystyle \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ where $\lfloor x \rfloor $ represent floor function of $x$.
My Try:: Here $\displaystyle f(x) = \frac{x^2}{\sin x\cdot \tan x}$ is an even function.
So we will calculate for $\displaystyle \lim_{x\to 0^{+}}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$
Put $x=0+h$ where $h$ is a small positive quantity, and using series expansion
So limit convert into $\displaystyle \lim_{h\to 0}\left\lfloor \frac{h^2}{\sin h\cdot \tan h}\right\rfloor = \lim_{h\to 0}\left\lfloor \dfrac{h^2}{\left(h-\dfrac{h^3}{3!}+\dfrac{h^5}{5!}- \cdots\right)\cdot \left(h+\dfrac{h^3}{3}+\dfrac{2}{15}h^5+ \cdot\right)}\right\rfloor$
Now how can i solve after that, Help me
Thanks
| Hint?
$$
\lim_{x\to 0} \frac{x^2}{\sin x \cdot \tan x}
= \lim_{x\to 0} \cos x \left( \frac{x}{\sin x} \right)^2
$$
Now we can split $\lim ab=\lim a \cdot \lim b$, since both limits exists. Alternatively one can use L'ôpitals rule twice. $\lim_{x\to a} f/g = \lim_{x\to a} f'/g'$. Somewhat of a hazzle but it works
$$
\lim_{x\to 0} \frac{x^2}{\sin x \cdot \tan x}
= \lim_{x\to 0} \frac{2x}{\cos x \tan x + \sin x \sec^2x}
= \lim_{x\to 0} \frac{2}{\cos x + 1/\cos x + 2 \sin^2x \sec^2x}
$$
The rest now follows from $\sin 0 = 0$ and $\cos 0 = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$x^3-3x^2+4x-2$ cannot be factored over $\mathbb R$ I'm new to the site, and I need a bit of help from you.
How can I prove that the polynomial: $f(x)=x^3-3x^2+4x-2$
cannot be factored as a product of polynomials of degree 1 with real coefficients?
Thanks.
| Observe that$$\begin{align*}
f(x)=x^3-3x^2+4x-2&=(x^3-3x^2+3x-1)+(x-1)\\\\
&=(x-1)^3+(x-1)\\\\
&=(x-1)((x-1)^2+1)\\\\
&=(x-1)(x^2-2x+2)
\end{align*}$$
By the quadratic formula, the roots of $x^2-2x+2$ are complex, so $f$ cannot be factored any further into polynomials with real coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find $\lim_{x \to 0^-}\frac{\sqrt{1 - \cos4x}}{\sin3x - \sin x}$ Find $ \lim_{x \to 0^-}\frac{\sqrt{1 - \cos4x}}{\sin3x - \sin x}$
What I did
$\frac{\sqrt{1 - \cos4x}}{\sin3x - \sin x}$
$\frac{\sqrt{2}\sin2x}{\sin3x - \sin x}$
$\frac{\sqrt{2}\sin2x}{2\cos 2x \sin x}$
$\frac{\sqrt{2}(2 \sin x \cos x)}{2\cos 2x \sin x}$
$\lim_{x \to 0^-}\frac{\sqrt{2}cosx}{\cos 2x} = \sqrt{2}$
But the answer is not $\sqrt{2}$
| $$\begin{gathered}
\mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {1 - \cos \left( {4x} \right)} }}
{{\sin \left( {3x} \right) - \sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {2{{\sin }^2}\left( {2x} \right)} }}
{{2\cos \left( {2x} \right)\sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt 2 \left| {\sin \left( {2x} \right)} \right|}}
{{2\cos \left( {2x} \right)\sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - \sqrt 2 \sin \left( {2x} \right)}}
{{2\cos \left( {2x} \right)\sin \left( x \right)}} \hfill \\
= \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - 2\sqrt 2 \sin \left( x \right)\cos \left( x \right)}}
{{2\cos \left( {2x} \right)\sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - 2\sqrt 2 \cos \left( x \right)}}
{{2\cos \left( {2x} \right)}} = - \sqrt 2 . \hfill \\
\end{gathered} $$
Note that $\sin \left( u \right) > 0,\,\,\forall u \in \left( {0,\frac{\pi }
{2}} \right) \Rightarrow \sin \left( u \right) < 0,\,\,\forall u \in \left( { - \frac{\pi }{2},0} \right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show $\sum_{n=1}^\infty\frac{1}{n^2+3n+1}=\frac{\pi\sqrt{5}}{5}\tan\frac{\pi\sqrt{5}}{2}$. How to show that
$$\sum_{n=1}^\infty\frac{1}{n^2+3n+1}=\frac{\pi\sqrt{5}}{5}\tan\frac{\pi\sqrt{5}}{2}$$
?
My try:
We have
$$n+3n+1=\left(n+\frac{3+\sqrt{5}}{2}\right)\left(n+\frac{3-\sqrt{5}}{2}\right),$$
so
$$\frac{1}{n^2+3n+1}=\frac{2}{\sqrt{5}}\left(\frac{1}{2n+3-\sqrt{5}}-\frac{1}{2n+3+\sqrt{5}}\right).$$
Then, I don't know how to proceed.
| Note
$$ n^2+3n+1=(n+\frac{3}{2})^2+\left(\frac{\sqrt 5i}{2}\right)^2 $$
and
and hence
\begin{eqnarray}
\sum_{n=0}^\infty\frac{1}{n^2+3n+1}&=&\sum_{n=0}^\infty\frac1{(n+\frac{3}{2})^2+\left(\frac{\sqrt 5i}{2}\right)^2}\\
&=&\frac12\sum_{n=-\infty}^\infty\frac1{(n+\frac{3}{2})^2+\left(\frac{\sqrt 5i}{2}\right)^2}.
\end{eqnarray}
Then using the result from this, you will get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find limit of $\frac {1}{x^2}- \frac {1}{\sin^2(x)}$ as x goes to 0 I need to use a taylor expansion to find the limit.
I combine the two terms into one, but I get limit of $\dfrac{\sin^2(x)-x^2}{x^2\sin^2(x)}$ as $x$ goes to $0$. I know what the taylor polynomial of $\sin(x)$ centered around $0$ is… but now what do I do?
| Using Taylor series
$$\lim_{x\rightarrow0} \frac{\sin^2 x - x^2}{x^2\sin^2 x} = \lim_{x\rightarrow0} \frac{\Big(x - \frac{x^3}{6} + O(x^5)\Big)^2 - x^2}{x^2\Big(x - \frac{x^3}{6} + O(x^5)\Big)^2} = \lim_{x\rightarrow 0} \frac{\Big(x^2 - \frac{x^4}{3} + O(x^6)\Big) - x^2}{x^2\Big(x^2 - \frac{x^4}{3} + O(x^6)\Big)} \\= \lim_{x\rightarrow 0} \frac{-\frac{x^4}{3} + O(x^6)}{x^4 + O(x^6)} = -\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the value of the expression $2x^2 + 3xy – 4y^2$ when $x = 2$ and $y = - 4$? What is the value of the expression $2x^2 + 3xy – 4y^2$ when $x = 2$ and $y = - 4$?
I'm not good at algebra so please explain in easy to understand steps.
Thanks
| $2x^2 + 3xy - 4y^2$ when $x = 2$ and $y = -4$,
$2(2)^2 + 3(2)(-4) - 4(-4)^2 = 2 *4 - 24 - 4*16 = 8 - 24 - 64 = -80$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
| You can easily see that $x^6 - 1$ has a root at $1$, so you know that
$$x^6 - 1 = (x-1) \cdot p(x)$$
Where $p(x)$ is a polynomial of degree $5$. Perform polynomial division to find $p$:
$$\begin{align*} (x^6 - 1) \div (x-1) & = x^5 + (x^5 - 1) \div (x-1) \\
& = x^5 + x^4 + (x^4 - 1) \div (x-1) \\
& = x^5 + x^4 + x^3 + (x^3 - 1) \div (x-1) \\
& = x^5 + x^4 + x^3 + x^2 + (x^2 - 1) \div (x-1) \\
& = x^5 + x^4 + x^3 + x^2 + x + (x-1) \div (x-1) \\
& = x^5 + x^4 + x^3 + x^2 + x + 1
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 12,
"answer_id": 3
} |
Find value of sum of reciprocals of powers of a number Is there a simple way to find the value of the following expression?
$$\frac1x+\frac1{x^2}+\frac1{x^3}+\cdots$$
On trial and error, I was getting $\frac1{x-1}$, but I'm looking for a mathematical proof to it.
Please don't use complicated notation like summation unless absolutely necessary, because I'm not too familiar with it.
Edit: I tried another method. Let the answer be $a$. If we calculate $ax$, we get what appears to be $1+a$.
$$ax=1+a$$
$$ax-a=1$$
$$a(x-1)=1$$
$$a=\frac1{x-1}$$
Is that sufficient to prove the answer?
| Define the $n$th partial sum by
$$
S_n = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots + \frac{1}{x^n}
$$
Then
\begin{align*}
x S_n &= 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots + \frac{1}{x^{n-1}} \\
\implies x S_n - S_n &= 1 - \frac{1}{x^n} \iff S_n = \frac{1- \frac{1}{x^n}}{x-1}, \; x \neq 1
\end{align*}
Now assume $\left| \frac{1}{x} \right| < 1$. The limit $n \to \infty$ is then
\begin{align*}
S := \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{1- \frac{1}{x^n}}{x-1} = \frac{1}{x-1},
\end{align*}
because $(1/x)^n$ goes to $0$ if $|1/x| < 1$.
If $|1/x| > 1$ the limit is $ \pm \infty$ and the sum diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Using the half/double angle formulas to solve an equation I am completely stumped by this problem: $$\cos\theta - \sin\theta =\sqrt{2} \sin\frac{\theta}{2} $$
I know that I should start by isolating $\sin\dfrac{\theta}{2}$
and end up with $$\frac {\cos\theta - \sin\theta}{\sqrt{2}} = \sin\frac{\theta}{ 2}$$
From here on out, I have no idea what steps to take.
| $$\cos \theta . \sin \frac{\pi}{4} - \cos \frac{\pi}{4} . \sin \theta = \sin \frac{\theta}{2}$$
$$\sin (\frac{\pi}{4} - \theta) = \sin \frac{\theta}{2}$$
$$\frac{\pi}{4} - \theta = n\pi + (-1)^n(\frac{\theta}{2})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integral with series How do I represent this integral
$$\int_{0}^{1} \frac{10}{10+x^4} dx$$
as a series so that I can calculate with an error of less than $10^{-5}$.
| \begin{align}
\frac{10}{10+x^4} = \frac{1}{1-\left( \frac{-x^4}{10} \right)} = \frac 1 {1-r} & = 1+r+r^2+r^3+\cdots \\[6pt]
& = 1-\frac{x^4}{10} + \frac{x^8}{100} - \cdots
\end{align}
Integrating term by term from $0$ to $1$ gives
$$
1 - \frac 1 {50} + \frac 1 {900} -\cdots
$$
Since the terms alternate in sign and get smaller, and approach $0$, the error after each term is always smaller in absolute value than the next term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How to sum $\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}$? Does anyone know the general strategy for summing a series of the form:
$$\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2},$$
where $a$ is a positive integer?
Any hints or ideas would be great!
| Certain series of the form $\sum_{n = -\infty}^\infty f(n)$ can be evaluated by means of residue calculus. One important result which is useful to this problem states that if $f$ is holomorphic on $\Bbb C \setminus \{z_1,\ldots, z_k\}$ (where the $z_i$ are the isolated singularities of $f$) and $|zf(z)|$ is bounded for $|z|$ sufficiently large, then $\sum_{n = -\infty}^\infty f(n)$ is the negative of the sum of the residues of $\pi \cot(\pi z)f(z)$ at the $z_i$, provided none of the $z_i$ are integers. You can find more information in Marsden's Basic Complex Analysis text.
To apply the result to this problem, first note that $$\sum_{n = 1}^\infty \frac{1}{n^2 + a^2} = \frac{1}{2}\sum_{n = -\infty}^\infty \frac{1}{n^2 + a^2} - \frac{1}{2a^2}$$ Let $f(z) = \frac{1}{z^2 + a^2}$. Then $f$ has simple poles at $z = ai$ and $z = -ai$. For $|z| \ge \max\{1, 2a\}$, $$|zf(z)| \le \frac{|z|}{|z|^2 - a^2} \le \dfrac{|z|}{|z|^2 - \frac{|z|^2}{4}} = \frac{4}{3|z|} \le \frac{4}{3}$$
Therefore
\begin{align}-\sum_{n = -\infty}^\infty \frac{1}{n^2 + a^2} &= \text{Res}_{z = ai} \frac{\pi\cot(\pi z)}{z^2 + a^2} + \text{Res}_{z = -ai} \frac{\pi\cot(\pi z)}{z^2 + a^2}\\
&= \frac{\pi\cot(\pi ai)}{2ai} - \frac{\pi\cot(-\pi ai)}{2ai}\\
&= -\frac{\pi\coth(\pi a)}{2a} - \frac{\pi \coth(\pi a)}{2a}\\
&= -\frac{\pi\coth(\pi a)}{a}\end{align}
Hence,
$$\sum_{n = 1}^\infty \frac{1}{n^2 + a^2} = \frac{\pi\coth(\pi a)}{2a} - \frac{1}{2a^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Solve cubic equation $x^3-9 x^2-15x-6 =0$ without going Cardano
Solve the cubic equation for $x\in\mathbb{R}$ $$x^3-9 x^2-15x-6 =0$$
Note that the only real solution is $x=3+2\sqrt[3]{7}+\sqrt[3]{7^2}$. Given the regularity of this solution, can we solve for it constructively, without going full Cardano?.
Also, can we prove that there is only one real solution without using the discriminant?
| When you say "full Cardano" are you referring to using a formula? Because there is a process for solving a cubic that is constructive. But maybe you do not want this because it is essentially how to construct Cardano's formula.
Substituting $x=y+3$ leads to:
$$\begin{align}
x^3-9x^2-15x-6&=0\\
(y+3)^3-9(y+3)^2-15(y+3)-6&=0\\
y^3-42y-105&=0
\end{align}$$
Now introduce parameters $u$ and $v$ such that their sum is a solution $y$:
$$\begin{align}
(u+v)^3-42(u+v)-105&=0\\
u^3+v^3+(3uv-42)(u+v)-105&=0
\end{align}$$
We have freedom to choose $u$ such that $3uv-42=0$. That implies $v=\frac{14}{u}$ and simplifies the equation above:
$$\begin{align}
u^3+\left(\frac{14}{u}\right)^3-105&=0\\
u^6-105u^3+14^3&=0
\end{align}$$
This is quadratic in $u^3$:
$$\begin{align}
u^3&=\frac{105\pm\sqrt{105^2-4\cdot14^3}}{2}\\
u^3&=\frac{105\pm7}{2}\\
\end{align}$$
You may assume the "$+$" for $u^3$, since the "$-$" would give the symmetric $v^3$.
$$\begin{align}
u^3&=\frac{105+7}{2}=56\\
u&=2\sqrt[3]{7}
\end{align}$$
From which $v=\frac{14}{2\sqrt[3]{7}}=\frac{7}{\sqrt[3]{7}}=\sqrt[3]{7^2}$.
And now $x=y+3=u+v+3=2\sqrt[3]{7}+\sqrt[3]{7^2}+3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Converting a polar equation to a rectangular one $$r=\frac { 4 }{ 1+2\sin\theta } $$
Steps I took:
$$(1+2\sin\theta )r=\frac { 4 }{ 1+2\sin\theta } (1+2\sin\theta )$$
$$r+2r\sin\theta =4$$
$$r+2y=4$$
$$(r+2y)^2=16$$
$$(r+2y)(r+2y)=r^2+4yr+4y^2$$
$$r^2+4yr+4y^2-16=0$$
My outcome doesn't seem to match the correct answer. Where did I go wrong?
| use that $$r=\sqrt{x^2+y^2}$$ and $$\sin(\theta)=\frac{y}{\sqrt{x^2+y^2}}$$ and you will get
$$\sqrt{x^2+y^2}=\frac{4}{1+\frac{2y}{\sqrt{x^2+y^2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $c_n = \frac1n \bigl(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \bigr)$ converges I want to show that $c_n$ converges to a value $L$ where:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n}$$
First, it's obvious that $c_n > 0$.
I was able to show using the following method that $c_n$ is bounded:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n} < \overbrace{\frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{2}}}{n}}^{n-1 \text{ times}} = \frac{\large {n - 1}}{n\sqrt{2}} < \frac{1}{\sqrt{2}}$$
So now we know that $\large {0 < c_n < \frac{1}{\sqrt{2}}}$.
I know from testing for large values of $n$ that $c_n \to 0$.
What's left is actually finding a way to show this.
Any hints?
| Just wondering that the following elementary way is not among the presented solutions.
So, I add it here as a late answer:
The following inequality for $k \in \mathbb{N}$ will be used:
$$\sqrt{k+1}-\sqrt k = \frac 1{\sqrt{k+1}+\sqrt k} > \frac 1{2\sqrt{k+1}}$$
Hence,
\begin{eqnarray*} \frac 1n\sum_{k=1}^n \frac 1{\sqrt k}
& = & \frac 1n\left(1+ \sum_{k=1}^{n-1} \frac 1{\sqrt{k+1}}\right) \\
& < & \frac 1n\left(1+ 2\sum_{k=1}^{n-1} (\sqrt{k+1}-\sqrt k)\right) \\
& = & \frac 1n\left(1+ 2(\sqrt{n}-1)\right) \\
& = & -\frac 1n + \frac 2{\sqrt n} \stackrel{n\to\infty}{\longrightarrow} 0
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
Prove that $\int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x = \frac{t^{n+1}}{n+1} + o(t^n)$, when $t \to \infty,\,n\in\Bbb{R}^+$ I hae to prove that
$$\int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x = \frac{t^{n+1}}{n+1} + o(t^n), \quad\text{ when } t \to \infty,\,n\in\Bbb{R}^+$$
where $o(\cdot)$ is the Little-o notation.
What I have done so far:
Let $F_n(t) = \int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x$, then $F_0(t) =\int_0^\infty\frac{1}{1+e^{x-t}}\mathrm{d}x = \log(e^t+1)$. The derivative of $F_n(t)$ is:
\begin{align*}
F_n'(t) &= \int_0^\infty\frac{e^{x-t}x^n}{(1+e^{x-t})^2}\mathrm{d}x\\
&= -\int_0^\infty x^n\mathrm{d}\frac{1}{1+e^{x-t}}\\
&= \left.\frac{x^n}{1+e^{x-t}}\right|_0^\infty+\int_0^\infty \frac{1}{1+e^{x-t}}\mathrm{d}x^n\\
&= \int_0^\infty \frac{nx^{n-1}}{1+e^{x-t}}\mathrm{d}x\\
&= nF_{n-1}(t)
\end{align*}
| First, let's split up the integral:
$$
\begin{align}
\int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x
&=\int_{-t}^\infty\frac{(x+t)^n}{1+e^x}\mathrm{d}x\\
&=\color{#C00000}{\int_{-t}^0\frac{(x+t)^n}{1+e^x}\mathrm{d}x}+\color{#00A000}{\int_0^\infty\frac{(x+t)^n}{1+e^x}\mathrm{d}x}\tag{1}
\end{align}
$$
Note that the first integral on the right side of $(1)$ is
$$
\begin{align}
\color{#C00000}{\int_{-t}^0\frac{(x+t)^n}{1+e^x}\mathrm{d}x}
&=t^{n+1}\int_{-1}^0\frac{(1+x)^n}{1+e^{tx}}\mathrm{d}x\\
&=t^{n+1}\int_{-1}^0(1+x)^n\,\mathrm{d}x-\color{#0000FF}{t^{n+1}\int_{-1}^0\frac{(1+x)^n}{1+e^{tx}}e^{tx}\,\mathrm{d}x}\\
&=\frac{t^{n+1}}{n+1}+O\left(t^n\right)\tag{2}
\end{align}
$$
because
$$
\begin{align}
\color{#0000FF}{t^{n+1}\int_{-1}^0\frac{(1+x)^n}{1+e^{tx}}e^{tx}\,\mathrm{d}x}
&\le t^{n+1}\int_{-1}^0(1+x)^n\,e^{tx}\mathrm{d}x\\
&\le t^{n+1}\int_{-1}^0e^{nx}\,e^{tx}\mathrm{d}x\\
&\le\frac{t^{n+1}}{n+t}\\[6pt]
&=O\left(t^n\right)\tag{3}
\end{align}
$$
Furthermore, by dominated convergence
$$
\lim_{t\to\infty}\int_0^\infty\left(1+\frac xt\right)^n\,e^{-x}\,\mathrm{d}x
=1\tag{4}
$$
therefore, the second integral on the right side of $(1)$ is
$$
\begin{align}
\color{#00A000}{\int_0^\infty\frac{(x+t)^n}{1+e^x}\mathrm{d}x}
&\le t^n\int_0^\infty\left(1+\frac xt\right)^n\,e^{-x}\,\mathrm{d}x\\
&=O\left(t^n\right)\tag{5}
\end{align}
$$
Combining $(1)$, $(2)$, and $(5)$, we get
$$
\int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x
=\frac{t^{n+1}}{n+1}+O\left(t^n\right)\tag{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $2x^2+5y^2+6xy-2x-4y+1=0$ in real numbers
Solve the equation $2x^2+5y^2+6xy-2x-4y+1=0$
The problem does not say it but I think solutions should be from $\mathbb{R}$. I tried to express the left sum as a sum of squares but that does not work out. Any suggestions?
| $$2x^2 + 5y^2 + 6xy -2x -4y+1=0$$
$$(1+1)x^2 + (4+1)y^2 + (4+2)xy - 2x -4y + 1=0$$
$$(x^2 +4y^2 +4xy -2x -4y + 1) + (x^2 +2xy + y^2)=0$$
$$(x+2y-1)^2 + (x+y)^2=0$$
Your idea to write the expression as a sum of squares is good intuition. Now, when can a sum of squares be zero? Exactly when both of the squares are zero.
Thus
$$x+2y-1=0$$
$$x+y=0$$
You should be able to find the solution from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove $4^k - 1$ is divisible by $3$ for $k = 1, 2, 3, \dots$ For example:
$$\begin{align}
4^{1} - 1 \mod 3 &=
\\
4 -1 \mod 3 &=
\\
3 \mod 3 &=
\\3*1 \mod 3 &=0
\\
\\
4^{2} - 1 \mod 3 &=
\\
16 -1 \mod 3 &=
\\
15 \mod 3 &=
\\3*5 \mod 3 &= 0
\\
\\
4^{3} - 1 \mod 3 &=
\\
64 -1 \mod 3 &=
\\
21 \mod 3 &=
\\3*7 \mod 3 &=
0\end{align}
$$
Define $x = \frac{4^k - 1}{3}$. So far I have:
$$k_1 \to 1 \Longrightarrow x_1 \to 1
\\
k_2 \to 2 \Longrightarrow x_2 \to 5
\\
k_3 \to 3 \Longrightarrow x_3 \to 21
\\
k_4 \to 4 \Longrightarrow x_4 \to 85$$
But then it's evident that
$$4^{k_n} = x_{n+1} - x_n$$
I don't know if this helps, these are ideas floating in my head.
| You may also go about this easily with induction:
$$4^{n+1}-1 = (3+1)\cdot 4^n-1 = 3\cdot 4^n + (4^{n}-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 0
} |
Least squares with matrix in $GF(2)$? Here's an example of a problem I'm working on involving finding combination of bit vectors that yield a certain sum (in the $GF(2)$ sense):
$
\begin{pmatrix}
1 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \\
1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\
1 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 \\
0 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 1 \\
1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\
0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \\ x_7 \\ x_8 \\ x_9 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 1 \\ 1 \\ 1 \end{pmatrix}$
An obvious answer is just the vector where $x_7$ is 1 with all other elements being zero. However, running simple gaussian elimination on the above matrix quickly leads you to the non-regular matrix:
$\begin{pmatrix}
1 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 1 & 0 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}$
Which, continuing to solve from there leads you to the (correct, but sub-optimal) solution:
$x^T = \begin{pmatrix}0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0\end{pmatrix}$
Is there a good method for solvings systems of equations in $GF(2)$ s.t. the answer is minimal in the $L_1$ norm sense?
| The norm you're talking about is the Hamming weight of an element and you want the element of minimum Hamming weight in your solution space. As far as I know there does not exist a good general algorithm for computing it.
By computing the nullspace of that matrix and then a reduced basis of that nullspace, you can probably eyeball how to get from your non-optimal solution to the minimum weight solution with not too much trial and error. I think that's the best procedure you can hope for save throwing a computer at it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why the differentiation of $e^x$ is $e^x?$ $$\frac{d}{dx} e^x=e^x$$
Please explain simply as I haven't studied the first principle of differentiation yet, but I know the basics of differentiation.
| \begin{align}
\frac{d}{dx}e^x &= \frac{d}{dx}\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \right)\\
&=0+\frac{1}{1!}+\frac{2x}{2!}+\frac{3x^2}{3!}+\dots\\
&=0+1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots\\
&=0 + e^x\\
&=e^x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 5
} |
Integration by nonobvious substitutions The standard technique for evaluating the integral
$$\int \sec x \,dx$$
is making the nonobvious substitution
$$u = \sec x + \tan x, \qquad du = (\sec x \tan x + \sec^2 x) dx,$$ which transforms the integral to
$$\int \frac{du}{u} .$$
What are other integrals that are nicely handled with nonobvious substitutions?
| If you're unaware of a substitution, which usually happens when you have'nt attempted a problem of that particular type, the substitution usually seems non-obvious(happened to me a lot when beginning to learn integration), some of the examples which I liked the most are:
From here:$$\int \frac{\mathrm{dx}}{x^4[x(x^5-1)]^{1/3}}$$
This one's by me: Let $x^5z^3=x^5-1$. So
$$x^5(z^3-1)=1\implies 5x^4(z^3-1)\mathrm{d}x+x^5(-3z^2\mathrm{d}z)=0\implies \mathrm{d}x=\frac{3xz^2\mathrm{d}z}{5(z^3-1)}$$
So:
$$\int \frac{\mathrm{d}x}{x^4[x(x^5-1)]^{1/3}}\text{ or }\int x^{-13/3}(x^5-1)^{-1/3}\mathrm{d}x\\
=\int x^{-13/3}(x^5z^3)^{-1/3}.\frac{3xz^2\mathrm{d}z}{5(z^3-1)}=\frac35\int \frac{x^{-13/3}x^{-5/3}z^{-1}xz^2\mathrm{d}z}{x^{-5}}\\
=\frac35\int z\;\mathrm{d}z=\frac{3}{10}\left(\frac{x^5-1}{x^5}\right)^{2/3}+\mathcal{C}$$
Another One:
From here $$K=\int\frac{\ln x\,dx}{x^2+2x+4}$$
Try the substitution $x = \dfrac{4}{y}$ to get:
This one's by JimmyK4542 $$I = \int_{0}^{\infty}\dfrac{\ln x}{x^2+2x+4}\,dx = \int_{-\infty}^{0}\dfrac{\ln \frac{4}{y}}{\frac{16}{y^2}+\frac{8}{y}+4} \cdot \dfrac{-4}{y^2}\,dy$$
$$= \int_{0}^{\infty}\dfrac{\ln 4 - \ln y}{y^2+2y+4}\,dy = \ln 4 \int_{0}^{\infty}\dfrac{\,dy}{y^2+2y+4} - I$$
Thus, $$2I = \ln 4 \int_{0}^{\infty}\dfrac{\,dx}{x^2+2x+4}$$, which is easy to compute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Number of Solutions of $y^2-6y+2x^2+8x=367$? Find the number of solutions in integers to the equation
$$y^2-6y+2x^2+8x=367$$
How should I go about solving this?
Thanks!
| Use the completing the square technique to get $(y-3)^2+2(x+2)^2 = 384$.
Make the substitution $a = y-3$ and $b = x+2$ to get $a^2+2b^2 = 384$.
Since $2b^2$ and $384$ are even, $a^2$ must be even, and hence $a$ is even. So let $a = 2a_1$.
Then, we get $4a_1^2+2b^2 = 384$, i.e. $2a_1^2+b^2 = 192$.
Since $2a_1^2$ and $192$ are even, $b^2$ must be even, and hence $b$ is even. So let $b = 2b_1$.
You can keep doing this until you get down to $2a_4^2+b_3^2 = 3$.
Then, its easy to find all solutions and undo all the substitutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Inequality involving $\frac{\sin x}{x}$ Can anybody explain me, why the following inequality is true?
$$\sum_{k=0}^{\infty} \int_{k \pi + \frac{\pi}{4}}^{(k+1)\pi-\frac{\pi}{4}} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi - \frac{\pi}{4} \right)\right|}{(k+1) \pi - \frac{\pi}{4}} \ \frac{\pi}{2} $$
The question is motivated from the following calculation
$$
\sum_{k=0}^{\infty} \int_{k \pi}^{(k+1) \pi} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \sum_{k=0}^{\infty} \int_{k \pi + \frac{\pi}{4}}^{(k+1)\pi-\frac{\pi}{4}} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \\
\geq \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi - \frac{\pi}{4} \right)\right|}{(k+1) \pi - \frac{\pi}{4}} \ \frac{\pi}{2} = \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi - \frac{\pi}{4} \right)\right|}{2(k+1) - \frac{1}{2}} = \\
= \sum_{k=0}^{\infty} \frac{\sqrt{2}}{2} \frac{1}{2(k+1) - \frac{1}{2}} = \sum_{k=0}^{\infty} \frac{\sqrt{2}}{4k+3} \geq \sum_{k=0}^{\infty} \frac{\sqrt{2}}{4k} = \\
= \frac{\sqrt{2}}{4}\sum_{k=0}^{\infty} \frac{1}{k} = \infty
$$
which shows that $\frac{\sin \xi}{\xi} \notin \mathcal{L}^1(\mathbb{R})$.
| Hint
For $x\in[k\pi+\pi/4,(k+1)\pi-\pi/4]$,
$$|\sin x|\ge|\sin\big((k+1)\pi-\pi/4\big)|$$
and
$$x\le(k+1)\pi-\pi/4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the greatest common divisor of $2003^4 + 1$ and $2003^3 + 1$ Find the greatest common divisor of $2003^4 + 1$ and $2003^3 + 1$ without the use of a calculator. It is clear that $2003^4+1$ has a $082$ at the end of its number so $2003^4+1$ only has one factor of 2, while $2003^3+1$ has a $028$ at the end of its number so $2003^3+1$ has 2 factors of 2. The answer is supposed to be 2, but at this point it could be greater than 2. Any ideas? This was on a previous qualifying exam at BU.
| $d = \text{gcd}(2003^4+1,2003^3+1) \Rightarrow d \mid 2003^4-2003^3 = 2003^3\cdot 2002 = 2003^3\cdot 2\cdot 7\cdot 11\cdot 13$, and by checking case by case, we have $2$ is the only number such that $d$ divides into evenly. So $d = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$? Plotting both functions $\ln(1+\frac{1}{x-1})$ and $\frac{1}{x}$ in $[2,\infty)$ gives the impression that $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$.
Is it possible to prove it?
| We know that for $|x| < 1$, $$\log(1-x) = -x-\frac{1}{2}x^2 - \frac{1}{3}x^3 - \dotsb.$$
Since $$1 + \frac{1}{x-1} = \frac{x-1+1}{x-1} = \frac{x}{x-1},$$
we see that
$$\log \left(1+\frac{1}{x-1}\right) = \log \left(\frac{x}{x-1}\right) = -\log\left(\frac{x-1}{x}\right) = -\log\left(1-\frac{1}{x}\right).$$
So if $x \ge 2$ then $0 < \frac{1}{x} < 1$, so we can apply the power series:
$$-\log\left(1-\frac{1}{x}\right) = \frac{1}{x} + \frac{1}{2x^2} + \frac{1}{3x^3} + \dotsb > \frac{1}{x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the zeros of $f(x)=x^3+64$ $$f(x)=x^3+64$$
Again, I am really not sure how to do this I tried to factor but it clearly was not the right answer
| $x^3 + 64 = x^3+4^3 = (x+4)(x^2-4x+16) = (x+4)((x-2)^2+12) = 0 \to x+4 = 0 \to x= -4$. I suppose you want real roots. If you want complex roots, then $(x-2)^2 = -12 = (2\sqrt{3}\cdot i)^2 \to x = 2 \pm 2\sqrt{3}\cdot i$.
In general, if you are interested in finding the roots of $p(x) = x^3 + q$, then you first write: $p(x) = x^3 + (\sqrt[3]{q})^3 = (x+\sqrt[3]{q})(\cdots)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Getting wrong answer trying to evaluate $\int \frac {\sin(2x)dx}{(1+\cos(2x))^2}$ I'm trying to evaluate $$\int \frac {\sin(2x)dx}{(1+\cos(2x))^2} = I$$
Here's what I've got:
$$t=1+\cos(2x)$$
$$dt=-2\sin(2x)dx \implies dx = \frac {dt}{-2}$$
$$I = \int \frac 1{t^2} \cdot \frac {-1}2 dt = -\frac 12 \int t^{-2}dt = -\frac 12 \frac {t^{-2+1}}{-2+1}+C$$
$$=-\frac 12 \frac {t^{-1}}{-1}+C = \frac 1{2t} +C = \frac 1{2(1+\cos(2x)}+C$$
I need to find what went wrong with my solution. I know the answer is correct but the line $dt=-2\sin(2x)dx$ concludes that $dx=\frac {dt}{-2}$ is false. But why? What should it be instead?
| Probably, you saw the solution as
$$\frac{\sec^2(x)}{4}+C$$
But note that
$$\begin{array}{rclcc}
\frac{1}{2}\frac{1}{1+\cos(2x)}&=&\frac{1}{2}\frac{1}{1+\cos^2(x)-\sin^2(x)}&/\,\,\cos(2x)=\cos^2(x)-\sin^2(x)\\
&=&\frac{1}{2}\frac{1}{1+\cos^2(x)-\sin^2(x)}&/\,\,1=\cos^2(x)+\sin^2(x)\\
&=&\frac{1}{2}\frac{1}{2\cos^2(x)}\\
&=&\frac{\sec^2(x)}{4}\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Minimum of $\newcommand{\b}[1]{\bigl(#1\bigr)} \newcommand{\f}{\frac} \b{\f3a-1}^2+\b{\f ab-1}^2+\b{\f bc-1}^2+(3c-1)^2$ The minimum of :
$$\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\f}{\frac}
\b{\f3a-1}^2+\b{\f ab-1}^2+\b{\f bc-1}^2+(3c-1)^2$$ where $0<a,b,c\le9$, is $p-q\sqrt{r}$; $p,q,r\in\mathbb Z$ and $q,r$ are co-primes, then $(p+q+r)$ is equal to?
I tried expand and AM-GM but what to do about negative terms?
| USe Cauchy-Schwarz inequality we have
$$x^2+y^2+z^2+w^2\ge\dfrac{1}{4}(x+y+z+w)^2$$
so
$$LHS\ge \dfrac{1}{4}\left(\dfrac{3}{a}+\dfrac{a}{b}+\dfrac{b}{c}+3c-4\right)^2$$
and Use AM-GM inequality we have
$$\dfrac{3}{a}+\dfrac{a}{b}+\dfrac{b}{c}+3c\ge 4\sqrt[4]{\dfrac{3}{a}\cdot\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot 3c}= 4\sqrt[4]{9}=4\sqrt{3}$$
so
$$LHS\ge 4(\sqrt{3}-1)^2=16-8\sqrt{3}$$
if and only if
$$\dfrac{3}{a}=\dfrac{a}{b}=\dfrac{b}{c}=3c=\sqrt{3}\Longrightarrow a=\sqrt{3},b=1,c=\dfrac{\sqrt{3}}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to define an affine transformation using 2 triangles? I have $2$ triangles ($6$ dots) on a $2D$ plane.
The points of the triangles are: a, b, c and x, y, z
I would like to find a matrix, using I can transform every point in the 2D space.
If I transform a, then the result is x. For b the result is y, and for c the result is z
And if there is a given d point, which is halfway from a to b, then after the transformation the result should be between x and y halfway.
I've tried to solve it according to NovaDenizen's solution, But the result is wrong.
The original triangle:
$$
a =
\left[
\begin{array}{ccc}
-3\\
0\\
\end{array}
\right]
$$
$$
b =
\left[
\begin{array}{ccc}
0\\
3\\
\end{array}
\right]
$$
$$
c =
\left[
\begin{array}{ccc}
3\\
0\\
\end{array}
\right]
$$
The x, y, z dots:
$$
x =
\left[
\begin{array}{ccc}
2\\
3\\
\end{array}
\right]
$$
$$
y =
\left[
\begin{array}{ccc}
3\\
2\\
\end{array}
\right]
$$
$$
z =
\left[
\begin{array}{ccc}
4\\
3\\
\end{array}
\right]
$$
I've created a figure:
I tried to transform the (0, 0) point, which is halfway between a and b, but the result was (3, 3.5) instead of (3, 3)
The T matrix is:
$$\left[
\begin{array}{ccc}
1/3 & 1/6 & 0\\
0 & -1/2 & 0\\
3 & 3,5 & 1\\
\end{array}
\right]$$
| The transformation you're looking for has this form:
$$\left[
\begin{array}{ccc}
t_1 & t_2 & t_3\\
t_4 & t_5 & t_6\\
0 & 0 & 1\\
\end{array}
\right]
\left[
\begin{array}{ccc}
a_1 & b_1 & c_1\\
a_2 & b_2 & c_2 \\
1 & 1 & 1 \\
\end{array}
\right]
= \left[\begin{array}{ccc}
x_1 & y_1 & z_1\\
x_2 & y_2 & z_2\\
1 & 1 & 1
\end{array}
\right]
$$
or
$${\bf T A} = {\bf X}$$
so
$${\bf T} = {\bf X}{\bf A}^{-1}$$
Now $\bf T$ is a transformation matrix you can use on any point, like
$${\bf T}\left[\begin{array}{c}a_1\\a_2\\1\end{array}\right] = \left[\begin{array}{c}x_1\\x_2\\1\end{array}\right]$$
It's linear, so it has the halfway point property you were looking for.
$$
{\bf A} = \left[\begin{array}{ccc}
-3 & 0 & 3\\
0 & 3 & 0\\
1 & 1 & 1\\
\end{array}\right]
$$
$$
\left[\begin{array}{ccc|ccc}
-3 & 0 & 3 & 1 & 0 & 0\\
0 & 3 & 0 & 0 & 1 & 0\\
1 & 1 & 1 & 0 & 0 & 1\\
\end{array}\right]
$$
$$
\left[\begin{array}{ccc|ccc}
1 & 0 & -1 & -\frac13 & 0 & 0\\
0 & 3 & 0 & 0 & 1 & 0\\
0 & 1 & 2 & \frac13 & 0 & 1\\
\end{array}\right]
$$
$$
\left[\begin{array}{ccc|ccc}
1 & 0 & -1 & -\frac13 & 0 & 0\\
0 & 1 & 0 & 0 & \frac13 & 0\\
0 & 0 & 2 & \frac13 & -\frac13 & 1\\
\end{array}\right]
$$
$$
\left[\begin{array}{ccc|ccc}
1 & 0 & 0 & -\frac16 & -\frac16 & \frac12\\
0 & 1 & 0 & 0 & \frac13 & 0\\
0 & 0 & 1 & \frac16 & -\frac16 & \frac12\\
\end{array}\right]\\
\text{so } {\bf A}^{-1} =
\left[\begin{array}{ccc}
-\frac16 & -\frac16 & \frac12\\
0 & \frac13 & 0\\
\frac16 & -\frac16 & \frac12\\
\end{array}\right]$$
$${\bf XA}^{-1} =
\left[\begin{array}{ccc}
2 & 3 & 4\\
3 & 2 & 3\\
1 & 1 & 1\\
\end{array}\right]
\left[\begin{array}{ccc}
-\frac16 & -\frac16 & \frac12\\
0 & \frac13 & 0\\
\frac16 & -\frac16 & \frac12\\
\end{array}\right]
=
\left[\begin{array}{ccc}
\frac13 & 0 & 3\\
0 & -\frac13 & 3\\
0 & 0 & 1\\
\end{array}\right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Show that: $\frac{1}{1+\log_ab+\log_bc}+\frac{1}{1+\log_bc+\log_ca}+\frac{1}{1+\log_ca+\log_ab}\leq1$
Let $a, b, c>1$. Show that:
$$\frac{1}{1+\log_ab+log_bc}+\frac{1}{1+\log_bc+\log_ca}+\frac{1}{1+\log_ca+\log_ab}\leq1.$$
My attempt:
We noted $\log_bc=x, \log_ca=y, \log_ab=z$ with $xyz=1$ and we have reduced inequality at this time:$$2(x+y+z)\leq xy^2+x^2y+yz^2+y^2z+xz^2+x^2z.$$
Does anyone have idea how it goes? Thank You!
I found a simple continuation:
$$2(x+y+z)\leq xy^2+x^2y+yz^2+y^2z+xz^2+x^2z\iff$$
$$\iff2(x+y+z)\leq xy(x+y)+yz(y+z)+zx(z+x)\iff$$
$$\iff2(x+y+z)\leq (xy+yz+zx)(x+y+z)-3xyz\iff$$
$$\iff3\leq(x+y+z)(xy+yz+zx-2).$$
The last inequality follows easily using:
$$3\leq x+y+z$$
$$1=3(xyz)^{\frac{2}{3}} -2\leq xy+yz+zx-2.$$
These last inequality is obtained by applying AM-GM inequality.
| other solution:
since $xyz=1$,we only prove
$$\dfrac{1}{1+x+y}+\dfrac{1}{1+y+z}+\dfrac{1}{1+x+z}\le 1$$
let
$$x=\dfrac{a^2}{bc},y=\dfrac{b^2}{ac},z=\dfrac{c^2}{ab}$$
so
$$\Longrightarrow \dfrac{1}{1+y+z}=\dfrac{1}{1+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}}=\dfrac{abc}{abc+b^3+c^3}$$
so this inequality we only prove
$$\Longleftrightarrow \dfrac{abc}{abc+a^3+c^3}+\dfrac{abc}{abc+b^3+c^3}+\dfrac{abc}{abc+a^3+b^3}\le 1$$
note
$$a^3+b^3\ge ab(a+b)\Longrightarrow a^3+b^3+abc\ge ab(a+b+c)$$
$$\Longrightarrow \dfrac{abc}{a^3+b^3+abc}\le\dfrac{abc}{ab(a+b+c)}=\dfrac{c}{a+b+c}$$
so
$$\sum_{cyc}\dfrac{abc}{a^3+b^3+abc}\le\sum_{cyc}\dfrac{c}{a+b+c}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
find the limit of the following expression Let $f\left(x\right)=\left(x^{2}+1\right)e^{x}$.
Find the following:$$\lim_{n\to+\infty}n\int_{0}^{1}\left(f\left(\frac{x^{2}}{n}\right)-1\right)\,dx$$
I dont know if L'Hopital's Rule may be used and tried using Integral Mean Value Theorem, but I cannot solve it! Any help?
| Explicitly,
$$
\lim_{n\to\infty} n \int_0^1 \left[f\left(\frac{x^2}{n}\right) -1\right]\,dx
= \lim_{n\to\infty} n \int_0^1 \left[\left(\frac{x^4}{n^2} + 1\right)e^{x^2/n} - 1\right]\,dx
$$
You can expand the exponential as a Taylor series:
\begin{align*}
n\int_0^1 \left[\left(\frac{x^4}{n^2} + 1\right)e^{x^2/n} - 1\right]\,dx
&= n\int_0^1 \Biggl[\frac{x^4}{n^2}\left\{1 + \frac{x^2}{n} + \frac{1}{2}\frac{x^4}{n^2} + \frac{1}{3!}\frac{x^6}{n^3}+\cdots\right\} \\
&\qquad\qquad+1\left\{\color{red}{1} + \color{green}{\frac{x^2}{n}} + \frac{1}{2}\frac{x^4}{n^2} + \frac{1}{3!}\frac{x^6}{n^3}+\cdots\right\} \\
&\qquad\qquad\color{red}{-1}\Biggr]\,dx\\
&= n\int_0^1\left[\color{green}{\frac{x^2}{n}}
+ \frac{3}{2}\frac{x^4}{n^2}
+ \frac{7}{6}\frac{x^6}{n^3} + \cdots
\right],dx
\end{align*}
Everything in the integrand after the first term is a multiple of $\frac{1}{n^2}$, so even after multiplying by $n$, the integrals will vanish as $n\to\infty$. So the limit is
$
\int_0^1 x^2\,dx = \frac{1}{3}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Trouble with definite integral calculating probabilities I cannot solve this:
$$\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \tan ^{-1}(a+\tan (x)) \, dx$$
it apeared when trying to find out the probability:
$$P\{\tan a - \tan b \leq 2x\},\ \ 0 < x < 1\sqrt{3}$$
Knowing that the joint distribution $f(a,b)$ is
$f(a,b) = \frac{2}{\pi^2}$ in the region $-\pi/2 < b < a < \pi/2$ (triangle)
"$a$" has marginal density:
$\frac{2x+\pi}{\pi^2}$ where $-\pi/2 < a < \pi/2$.
and "$b$" has density: $\frac{\pi-2x}{\pi^2}$ being $ -\pi/2 < b < a$.
| An approach that is less elegant but more elementary than residues.
Let
$$
J(b) = \int_{-\pi/2}^{\pi/2} \tan^{-1} \left( 2b + \tan x \right) dx.
$$
Then $J(0) = 0$ and using the substitution $\tan x = t$,
$$
\begin{align}
J'(b) &= \int_{-\infty}^\infty \frac{2}{(1+t^2)(1+(t+2b)^2)} dt
= \int_{-\infty}^\infty \frac{dt}{b(1+b^2)}\left[ \frac{3 b+t}{4 b^2+4 b t+t^2+1}+\frac{b-t}{t^2+1}\right]
\\&= \left.\frac{dt}{2b(1+b^2)}\left[ \log \left(\frac{4 b^2+4 b t+t^2+1}{t^2+1}\right)+2 b \tan ^{-1}(t)+2 b \tan ^{-1}(2 b+t)\right] \right\lvert_{-\infty}^\infty
\\&= \frac{\pi}{1+b^2}.
\end{align}
$$
By integrating back,
$$
J(b) = \int_0^b \frac{\pi \,db'}{1+b'^2} = \pi \tan^{-1} b.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$ If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$
I tried Schturm's method for quite some time, and Cauchy Schwarz for numerators because of the given product condition.
| We make the usual substitution $x = \dfrac{b}{a},y = \dfrac{c}{b}$ and $z = \dfrac{a}{c}$, then the inequality translates into proving:
$$\displaystyle 2\left(\sum\limits_{cyc} \frac{a}{b} - 3\right) \ge \sum\limits_{cyc} \frac{b}{a} - 3$$
Wlog assume $c$ lies between $a$ and $b$ (that is $a \le c \le b$ or $b \le c \le a$)
We have $\displaystyle \sum\limits_{cyc} \frac{a}{b} - 3 = \frac{(a-b)^2}{ab} - \frac{(a-c)(c-b)}{ac}$
and $\displaystyle \sum\limits_{cyc} \frac{b}{a} - 3 = \frac{(a-b)^2}{ab} - \frac{(a-c)(c-b)}{bc}$
Thus we are required to show: $\displaystyle \frac{(a-b)^2}{ab} \ge (a-c)(c-b)\left(\frac{2}{ac} - \frac{1}{bc}\right)$
Since, $(a-b)^2 = (a-c + c-b)^2 \ge 4(a-c)(c-b) \ge 0$ it suffices to prove
$\displaystyle \frac{4}{ab}+\frac{1}{bc} \ge \frac{2}{ac}$ which is true since $\displaystyle \frac{4}{ab} \ge \frac{2}{ac} \iff 2c \ge a \iff z \ge \frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Factor the polynomial $z^5 + 32$ in real factors The question that I have trouble solving is the following:
Factor the polynomial $z^5 + 32$ in real factors. The answer should not use trigonometric functions. (Hint: you are allowed to use the fact that: $cos(\pi/5) = \frac{1+\sqrt{5}}{4}$ and $cos(3\pi/5) = \frac{1-\sqrt{5}}{4}$.
In the previous question you were supposed to find the complex roots to the equation $z^5 = -32$ in polar form, resulting in the roots:
$$z_1 = 2e^{i\frac{\pi}{5}}$$$$z_2 = 2e^{i\frac{3\pi}{5}}$$ $$z_3 = 2e^{i\pi} = -2$$ $$z_4 = 2e^{i\frac{7\pi}{5}}$$$$ z_5 = 2e^{i\frac{9\pi}{5}} $$
My attempt at a solution:
First we know that there were only one real root to the equation $z^5 = -32$, namely $-2$.
So if we write the polynomial as the equation: $z^5 + 32 = 0$ we know that $(z+2)$ must be a factor and we have four potential factors left to find. Since we know that the polynomial $z^5 +32$ only has real coefficients, the non real roots must come in pairs.
We should then find two roots that are composed of conjugates of the roots with imaginary components to produce the other two real roots... But I am stuck here.
| $z^2 + 32 = (z+2)(x - \alpha)(z - \overline {\alpha})(z - \beta)(z- \overline {\beta})$ where $\overline {\alpha}, \overline {\beta}$ are complex conjugate of $\alpha, \beta$ respectively. Also $(z - \alpha)(z - \overline {\alpha})$ and $(z - \beta)(z- \overline {\beta})$ are real polynomials (why?). Now choose $\alpha = z_1, \beta = z_2.$ Then $\overline {\alpha} = z_5, \overline {\beta} = z_4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Sum of consecutive numbers I was wondering if there a way to figure out the number of ways to express an integer by adding up consecutive natural numbers.
For example for $N=3$ there is one way to express it
$1+2 = 3$
I have no idea where to start so any help would be appreciated
| The sum of the integers from $1$ to $n$ is $\dfrac{n(n+1)}{2}$. Hence, the sum of the integers from $m+1$ to $n$ is simply $\dfrac{n(n+1)}{2}-\dfrac{m(m+1)}{2}$. So, if the sum of the integers from $m+1$ to $n$ is $N$, then
$\dfrac{n(n+1)}{2}-\dfrac{m(m+1)}{2} = N$
$(n^2+n) - (m^2+m) = 2N$
$(n-m)(n+m+1) = 2N$
Hence, $n-m$ and $n+m+1$ are complementary factors of $2N$. Clearly, $n-m$ is smaller than $n+m+1$, and since $(n+m+1)-(n-m) = 2m+1$, the factors have opposite pairity.
For any $f_1$ and $f_2$ such that $2N = f_1f_2$, $f_1 > f_2$ and $f_1$, $f_2$ have opposite parity, we can solve $n+m+1 = f_1$ and $n-m = f_2$ to get $n = \dfrac{f_1+f_2-1}{2}$ and $m = \dfrac{f_1-f_2-1}{2}$.
Therefore, the number of ways to write $N = (m+1)+(m+2)+\cdots+(n-1)+n$ is simply the number of ways to factor $2N$ into two distinct positive integers with opposite parity.
Suppose $2N = 2^{k_0+1}p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$ where $p_1,p_2,\ldots,p_r$ are distinct odd primes. There are $(k_1+1)(k_2+1)\cdots(k_r+1)$ ways to divide the odd primes between $f_1$ and $f_2$. There are $2$ ways to give all $2$'s to one of the factors $f_1$, $f_2$. However, we need to divide by $2$ since this overcounts cases in which $f_1 < f_2$. Also, we need to subtract out the one trivial solution $n = N$ and $m = N-1$. This leaves us with $(k_1+1)(k_2+1)\cdots(k_r+1)-1$ ways to factor $2N$ into two distinct positive integers with opposite parity.
Therefore, if $N$ has prime factorization $N = 2^{k_0}p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$, then there are $(k_1+1)(k_2+1)\cdots(k_r+1)-1$ ways to write $N$ as the sum of two or more consecutive positive integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Prove the following inequality without using differentiation Let $a,b,c$ be real numbers that satisfy $0\le a,b,c\le 1$. Show that
$$\frac a{b+c+1} + \frac b{a+c+1} + \frac c{a+b+1} + (1-a)(1-b)(1-c) \le 1.$$
I don't know where to start. Multiplying everything by the denominators creates extreme mess.
| Let $f(a, b, c)$ denote the left hand side of the inequality. Since $$\frac{\partial^2}{\partial a^2}f=\frac{2b}{(a+c+1)^3}+\frac{2c}{(a+b+1)^3}\ge0$$we have that $f$ is convex in each of the three variables; hence,
the maximum must occur where $a, b, c \in \{0, 1\}$. Since $f$ is $1$ at each of these $8$ points,
the inequality follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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How to prove $\int_0^1 \frac1{1+x^2}\arctan\sqrt{\frac{1-x^2}2}d x=\pi^2/24$? Since I'm stuck at this final step of the solution here. I wished to try contour integral, taking the contour a quadrant with centre ($0$) and two finite end points of arc at $(1),(i)$:
Then:
$$\operatorname{Res}\limits_{x=i}\frac1{1+x^2}\arctan\sqrt{\frac{1-x^2}2}=-i\pi/8$$
But then:
$$2\pi i(-i\pi/8)=\pi^2/4??$$
| Disclaimer: now I have a working real-analytic technique, but it is quite a tour-de-force.
We have:
$$\begin{eqnarray*} I &=& \int_{0}^{1}\frac{1}{1+x^2}\arctan\sqrt{\frac{1-x^2}{2}}\,dx = \frac{1}{2}\int_{0}^{1}\frac{1}{(1+x)\sqrt{x}}\arctan\sqrt{\frac{1-x}{2}}\,dx\\&=&\int_{0}^{1/2}\frac{1}{(2-2x)\sqrt{1-2x}}\arctan\sqrt{x}\,dx=\int_{0}^{1/\sqrt{2}}\frac{x\arctan x}{(1-x^2)\sqrt{1-2x^2}}\,dx\end{eqnarray*}$$
and integrating by parts we get:
$$ I = \sqrt{2}\int_{0}^{1}\frac{\arctan\sqrt{1-x^2}}{2+x^2}\,dx=\sqrt{2}\int_{0}^{\pi/2}\frac{\cos\theta\arctan\cos\theta}{3-\cos^2\theta}d\theta.\tag{1}$$
so:
$$ I = \frac{1}{2\sqrt{2}}\int_{-\pi}^{\pi}\frac{\cos\theta\arctan\cos\theta}{3-\cos^2\theta}\,d\theta=\frac{1}{2\sqrt{2}}\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_{-\pi}^{\pi}\frac{\cos^{2n+2}\theta}{3-\cos^2\theta}\,d\theta.\tag{2}$$
On the other hand,
$$ I_n=\int_{-\pi}^{\pi}\frac{\cos^{2n+2}\theta}{3-\cos^2\theta}\,d\theta=-\int_{-\pi}^{\pi}\cos^{2n}\theta\,d\theta+3 I_{n-1}=-\frac{2\pi}{4^n}\binom{2n}{n}+3I_{n-1}\tag{3}$$
and $I_0=\pi\sqrt{\frac23}$. Now the plan is to solve recursion $(3)$ and compute the integral via $(2)$.
$$ I = \frac{1}{2\sqrt{2}}\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\sum_{m\geq 1}\frac{2\pi}{4^{n+m} 3^m}\binom{2n+2m}{n+m}\tag{4}$$
leads to:
$$ I = \frac{\pi}{\sqrt{2}}\sum_{n\geq 0}\sum_{m\geq 1}\frac{(-1)^{m}}{3^m(2n+1)}\binom{-1/2}{n+m}=\frac{\pi}{\sqrt{2}}\int_{0}^{1}\sum_{n\geq 0}\sum_{m\geq 1}\frac{(-1)^{m}x^{2n}\,dx}{3^m}\binom{-1/2}{n+m}\tag{5}$$
but since:
$$\sum_{n=0}^{s-1}(-1/3)^{s-n}(x^2)^n= \frac{1}{1+3x^2}\left((-1/3)^s-x^{2s}\right)$$
we have:
$$ I = \frac{\pi}{\sqrt{2}}\int_{0}^{1}\frac{dx}{1+3x^2}\sum_{s=1}^{+\infty}\left((-1/3)^s-x^{2s}\right)\binom{-1/2}{s}\tag{6}$$
and finally:
$$ I = \frac{\pi}{\sqrt{2}}\int_{0}^{1}\left(\sqrt{\frac{3}{2}}-\frac{1}{\sqrt{1+x^2}}\right)\frac{dx}{1+3x^2}\tag{7}$$
is easy to handle and leads to $\color{red}{\frac{\pi^2}{24}}$ as wanted, since:
$$ \int\frac{dx}{(1+3x^2)\sqrt{1+x^2}}=\frac{1}{\sqrt{2}}\arctan\frac{x\sqrt{2}}{\sqrt{1+x^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
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} |
How do I solve this, first I have to factor $ 2x\over x-1$ + $ 3x +1\over x-1$ - $ 1 + 9x + 2x^2\over x^2-1$? I am doing calculus exercises but I'm in trouble with this
$$\frac{ 2x}{x-1} + \frac{3x +1}{ x-1} - \frac{1 + 9x + 2x^2}{x^2-1}$$
the solution is
$$ 3x\over x+1$$
The only advance that I have done is factor $ x^2-1$ = $( x-1)$ $ (x+1)$.
I do not know how can I factor $1 + 9x + 2x^2$, can someone please guide me in how to solve this exercise.
| The first two terms are already over a common denominator.
First simplify $$\frac {2x}{x-1}+\frac {3x+1}{x-1}-\frac {1+9x+2x^2}{x^2-1}=\frac {5x+1}{x-1}-\frac {1+9x+2x^2}{x^2-1}$$
Then put the first fraction over the same denominator as the second $$\frac {(5x+1)(x+1)}{x^2-1}-\frac {1+9x+2x^2}{x^2-1}=\frac {(5x+1)(x+1)-(1+9x+2x^2)}{x^2-1}$$
Now simplify the numerator
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding a particular solution to a differential equation what is the particular solution for the following differential equation?
$$D^3 (D^2+D+1)(D^2+1)(D^2-3D+2)y=x^3+\cos\left(\frac{\sqrt{3}}2x \right)+xe^{2x}+\cos(x)$$
I tried Undetermined Coefficients and it took so long to solve it,not to mention it was on an exam. I was wondering if there could be any faster and simpler solution.
| Because this is an equation with constant coefficients, this in principle can be attacked using a Laplace transform. Here, for the sake of simplicity, I will assume that $y(0)$ and the first eight derivatives of $y$ at $x=0$ are zero. Then the Laplace transform of $y(x)$, $Y(p)$, is
$$\begin{align} Y(p) &= \left [ \frac{6}{p^4}+\frac{p}{p^2+1}+\frac{p}{p^2+\frac{3}{4}}+\frac{1}{(p-2)^2}\right ] \frac1{p^3 (p^2+p+1) (p^2+1) (p-2) (p-1)}\\ &= \frac{8 p^9-28 p^8+39 p^7+3 p^6-68 p^5+141 p^4-168 p^3+186 p^2-72 p+72}{ p^7 (p-1) (p-2)^3 \left(p^2+1\right)^2 \left(4 p^2+3\right)(p^2+p+1)} \end{align}$$
Inverting this Laplace Transform, as one might expect, will be an extremely messy business but is indeed possible by using the definition of the inverse LT and the Residue Theorem. The poles, their orders, and their residues are as follows:
$$\begin{array} \\ \text{pole} & \text{order} & \text{residue} \\ 0 & 7 & \frac{1}{960} \left(4 x^6+12 x^5-90 x^4+300 x^3+1950 x^2+310 x+125\right) \\ (i, -i) & 2 & \frac{1}{500} ((25 x+502) \sin{x}+(75 x+761) \cos{x}) \\ 1 & 1 & -\frac{113}{84} e^x \\ 2 & 3 & \frac{ 1}{52136000} \left(93100 x^2-747460 x+1929133\right) e^{2 x} \\ \left ( e^{i 2 \pi/3},e^{i 4 \pi/3} \right ) & 1 & \frac{1}{13377} \left( 13045 \sqrt{3} \sin \left(\frac{\sqrt{3} x}{2}\right)-9883 \cos \left(\frac{\sqrt{3} x}{2}\right) \right) e^{-x/2} \\ \left ( i \frac{\sqrt{3}}{2},- i \frac{\sqrt{3}}{2}\right ) & 1 & \frac{512 }{15561} \left(12 \cos \left(\frac{\sqrt{3} x}{2}\right)-41 \sqrt{3} \sin \left(\frac{\sqrt{3} x}{2}\right)\right)\end{array} $$
$y(x)$ is simply the sum of the residues in the right column of the above table.
| {
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"url": "https://math.stackexchange.com/questions/1103754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int \frac{x+2}{x^3-x} dx$ I need to solve this integral but I get stuck, let me show what I did:
$$\int \frac{x+2}{x^3-x} dx$$
then:
$$\int \frac{x}{x^3-x} + \int \frac{2}{x^3-x}$$
$$\int \frac{x}{x(x^2-1)} + 2\int \frac{1}{x^3-x}$$
$$\int \frac{1}{x^2-1} + 2\int \frac{1}{x^3-x}$$
now I need to resolve one integral at the time so:
$$\int \frac{1}{x^2-1}$$ with x = t I have:
$$\int \frac{1}{t^2-1}$$
Now I have no idea about how to procede with this...any help?
| Use partial fractions
$$\frac{x+2}{x^3-x}=\frac{x+2}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}.$$
Solve for $A,B$ and $C$ and then integrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does $\int_0^\infty \sin(x^{2/3}) dx$ converges? My Try:
We substitute $y = x^{2/3}$. Therefore, $x = y^{3/2}$ and $\frac{dx}{dy} = \frac{2}{3}\frac{dy}{y^{1/3}}$
Hence, the integral after substitution is:
$$ \frac{3}{2} \int_0^\infty \sin(y)\sqrt{y} dy$$
Let's look at:
$$\int_0^\infty \left|\sin(y)\sqrt{y} \right| dy = \sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| \sqrt{y} dy \ge \sum_{n=0}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| dy \\= \sum_{n=1}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\sqrt{\sin(y)^2}$$
| By $y = x^{2/3}$, we get $$\int_0^\infty \sin(x^{2/3})dx = \frac{3}{2}\int_0^\infty \sin(y)\sqrt{y}dy$$
Consider the integral over the interval $[2n\pi, 2(n+1)\pi]$, we have
$$
\begin{align}
\int_{2n\pi}^{2(n+1)\pi} \sin(y)\sqrt{y}dy &= \int_{2n\pi}^{(2n+1)\pi}\sin(y)\sqrt{y}dy + \int_{(2n+1)\pi}^{(2n+2)\pi}\sin(y)\sqrt{y}dy \\
&=\int_{2n\pi}^{(2n+1)\pi}\sin(y)\sqrt{y}dy + \int_{2n\pi}^{(2n+1)\pi}\sin(y+\pi)\sqrt{y+\pi}dy\\
&=\int_{2n\pi}^{(2n+1)\pi}\sin(y)\sqrt{y}dy - \int_{2n\pi}^{(2n+1)\pi}\sin(y)\sqrt{y+\pi}dy\\
&=-\int_{2n\pi}^{(2n+1)\pi}\sin(y)\frac{\pi}{\sqrt{y} + \sqrt{y+\pi}}dy\\
&\le -\int_{2n\pi}^{(2n+1)\pi}\sin(y)\frac{\pi}{\sqrt{(2n+1)\pi} + \sqrt{(2n+2)\pi}}dy\\
&=-\frac{2\pi}{\sqrt{(2n+1)\pi} + \sqrt{(2n+2)\pi}}
\end{align}$$
then $$\sum_{n=1}^\infty\int_{2n\pi}^{2(n+1)\pi} \sin(y)\sqrt{y}dy \leq \sum_{n=1}^\infty -\frac{2\pi}{\sqrt{(2n+1)\pi} + \sqrt{(2n+2)\pi}} =-\infty$$
so this integral does not converge. To be more convinced, see @Kyson's comment below(so this integral oscillates between $+\infty$ and $-\infty$)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that for any positve real Prove that for any positive real numbers $x,y,z$ such that $xyz \geq 1$
$$\frac{x^5-x^2}{x^5+y^2+z^2} + \frac{y^5-y^2}{y^5+z^2+x^2} +\frac{z^5-z^2}{z^5+x^2+y^2} \geq 0.$$
This problem is from the $2005$ IMO competition.
| Since,$$\frac{x^5-x^2}{x^5+y^2+z^2} - \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{x^2(x^3-1)^2(y^2+z^2)}{x^3(x^2+y^2+z^2)(x^5+y^2+z^2)} \ge 0$$
Hence, it suffices to prove $\displaystyle \sum\limits_{cyc} \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} \ge 0$
$$\sum\limits_{cyc} \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{\sum\limits_{cyc} \left(x^2 - \frac{1}{x}\right)}{x^2+y^2+z^2} \ge \frac{\sum\limits_{cyc} \left(x^2 - yz\right)}{x^2+y^2+z^2} = \frac{\sum\limits_{cyc} \left(x-y\right)^2}{2(x^2+y^2+z^2)} \ge 0$$
Since, $xyz \ge 1$.
Here are a few more proofs of the inequality, possibly neater.
| {
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"url": "https://math.stackexchange.com/questions/1112319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What Did I Do Wrong When Solving For This 2nd Order Differential Equation? (answered myself) $$
\frac{y''}{y'}+y' = f(x)
$$
I set the following to be true:
$$
y = \sum_{n=0}^{\infty} a_n x^n
$$
$$
f(x) = \sum_{n=0}^{\infty} b_nx^n
$$
Therefore:
$$
y'' = y'(f(x)-y')
$$
$$
\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n = \sum_{n=0}^{\infty} [(n+1)a_{n+1} x^n] \times \sum_{n=0}^{\infty} [(b_n - (n+1)a_{n+1})x^n]
$$
$$
\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n = \sum_{n=0}^{\infty} [(b_n - (n+1)a_{n+1})x^n \sum_{k=0}^n (k+1)a_{k+1}]
$$
Therefore
$$
(n+1)(n+2)a_{n+2} = (b_n - (n+1)a_{n+1}) \sum_{k=0}^n (k+1)a_{k+1}
$$
$$
b_n = \frac{(n+2)(n+1)a_{n+2}}{\sum_{k=1}^{n+1} ka_k} + (n+1)a_{n+1}
$$
$$
y = \sum_{n=0}^{\infty} (\frac{(n+2)(n+1)a_{n+2}}{\sum_{k=1}^{n+1} ka_{k}} + (n+1)a_{n+1})x^n
$$
Now if i set $a_n = 1$
$$
y = \sum_{n=0}^{\infty} (n+1)x^n + \sum_{n=0}^\infty (\frac{(n+2)(n+1)}{\sum_{k=1}^{n+1} k}x^n
$$
$$
\sum_{k=1}^{n+1} k = \frac{(n+1)(n+2)}{2}
$$
$$
y = \frac{1}{(x-1)^2} + \sum_{n=0}^\infty 2x^n
$$
$$
\sum_{n=0}^\infty 2x^n = \frac{2}{1-x}
$$
Therefore
$$
f(x) = \frac{2x-1}{(x-1)^2}
$$
$$
y = \frac{1}{1-x}
$$
Therefore
$$
\frac{-2(1-x)^2}{(1-x)^3}+\frac{1}{(1-x)^2} = \frac{2x-1}{(1-x)^2}
$$
Which is true.
| Your problem is that
you took the terms involving $b_n$
out of the summation.
Here is my take.
I'll start at
$\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n = \sum_{n=0}^{\infty} [(n+1)a_{n+1} x^n] \times \sum_{n=0}^{\infty} [(b_n - (n+1)a_{n+1})x^n]$
but use different indices for the summations
on the right.
$\begin{array}\\
\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n
&= \sum_{i=0}^{\infty} (i+1)a_{i+1} x^i \times \sum_{j=0}^{\infty} (b_j - (j+1)a_{j+1})x^j\\
&= \sum_{i=0}^{\infty}\sum_{j=0}^{\infty} (i+1)a_{i+1} (b_j - (j+1)a_{j+1})x^{i+j}\\
&= \sum_{n=0}^{\infty}\sum_{j=0}^{n} (n-j+1)a_{n-j+1} (b_j - (j+1)a_{j+1})x^{n}\\
&= \sum_{n=0}^{\infty}x^{n}\sum_{j=0}^{n} (n-j+1)a_{n-j+1} (b_j - (j+1)a_{j+1})\\
\end{array}
$
From this,
$\begin{array}\
(n+1)(n+2)a_{n+2}
&=\sum_{j=0}^{n} (n-j+1)a_{n-j+1} (b_j - (j+1)a_{j+1})\\
&=\sum_{j=1}^{n+1} ja_{j} (b_{n-j+1} - (n-j+2)a_{n-j+2})\\
\end{array}
$
I'll leave it at this.
| {
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"url": "https://math.stackexchange.com/questions/1114089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
computation $ \prod_{n=1}^{\infty}(1+q^{2n})(1+q^{2(n-1)}) = 2 \prod_{n=1}^{\infty} (1+q^{2n})^2 $ I want to compute the following identity
$ \prod_{n=1}^{\infty}(1+q^{2n})(1+q^{2(n-1)}) = 2 \prod_{n=1}^{\infty} (1+q^{2n})^2 = \frac{1}{2} \prod_{n=1}^{\infty} (1+q^{2(n-1)})^2 $
Can anyone gives some explict procedure of this identity?
| $\prod_{n=1}^\infty{(1+q^{2 \cdot n})\cdot(1+q^{2\cdot(n-1)})} = $
$[\prod_{n=1}^\infty{(1+q^{2 \cdot n})}]\cdot [\prod_{n=1}^\infty{(1+q^{2 \cdot (n - 1)})}] = $
$[\prod_{n=1}^\infty{(1+q^{2 \cdot n})}] \cdot [\prod_{n=0}^\infty{(1+q^{2 \cdot n})}] = $
$[\prod_{n=1}^\infty{(1+q^{2 \cdot n})}] \cdot (1+q^0) \cdot [\prod_{n=1}^\infty{(1+q^{2 \cdot n})}] = $
$2 \cdot \prod_{n=1}^\infty{(1+q^{2 \cdot n})^2}$
It's the same as simplifying a telescoping sequence like $\sum_{n=1}^\infty{\frac{1}{n \cdot (n+1)}}$ but with multiplication instead of addition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
factor the following expression $25x^2 +5xy -6y^2$ How to factor
$$25x^2 +5xy -6y^2$$
I tried with $5x(5x+y)-6y^2$. I'm stuck here.
I can't continue.
| Forget about the $y$ for a second.
If you want to factor $25x^2 + 5x - 6$, look for two numbers whose product is $25 \cdot -6 = -150$ and whose sum is $5$. It takes just a moment to see that the two numbers are $15$ and $-10$. Then:
\begin{align*}
25 x^2 + 5x - 6 &= 25 x^2 + (15x - 10x) - 6 \\ &= (25 x^2 + 15x) - (10x + 6) \\ &= 5x(5x + 3) - 2(5x + 3) \\& = (5x-2)(5x+3).\end{align*}
With the $y$ included the process is nearly identical:
\begin{align*}
25 x^2 + 5xy - 6 y^2&= 25 x^2 + (15xy - 10xy) - 6y^2 \\ &= (25 x^2 + 15xy) - (10xy + 6y^2) \\ &= 5x(5x + 3y) - 2y(5x + 3y) \\& = (5x-2y)(5x+3y).\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 0
} |
Find a parametrization of the intersection curve between two surfaces in $\mathbb{R^3}$ $x^2+y^2+z^2=1$ and $x^2+y^2=x$. Find a parametrization of the intersection curve between two surfaces in $\mathbb{R}^3$
$$x^2+y^2+z^2=1$$ and $$x^2+y^2=x.$$
I know that $x^2+y^2+z^2=1$ is a sphere and that $x^2+y^2=x$ is a circular cylinder. Any help is greatly appreciated, thank you.
| A point of the intersection belongs to the cylinder. So $(x-1/2)^2+y^2=1/4$. Take for parameter the angle $\theta$ such that $x-1/2=1/2 \cos \theta$ and $y=1/2 \sin \theta$. This is just the angle of the cylindrical coordinates.
Then you have $x+z^2=1$ which implies $z^2=1/2-1/2 \cos \theta=\sin^2(\theta/2)$. You then get $z=\sin(\theta/2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Transformation of an equation How do you get from the left side to the right side in this equation?
$$\frac{1+\sqrt{5}}{2} + 1 =\left(\frac{1+\sqrt{5}}{2}\right)^2$$
| We have $$\left(\frac{1+\sqrt{5}}{2}\right)^2=\frac{1}{4}(1+\sqrt{5})^2=\frac{1}{4}(6+2\sqrt{5})=\frac{3+\sqrt{5}}{2}= \frac{1+\sqrt{5}}{2}+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of the Determinant If $a^2+b^2+c^2+ab+bc+ca \le 0\quad \forall a, b, c\in\mathbb{R}$,
then find the value of the determinant
$$ \begin{vmatrix}
(a+b+2)^2 & a^2+b^2 & 1 \\
1 & (b+c+2)^2 & b^2+c^2 \\
c^2+a^2 & 1 & (c+a+2)^2 \\
\end{vmatrix}$$
I tried expanding the whole squares and using the identity $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$, but the result is incorrect. How should I evaluate this ?
| Hint
$$\dfrac{1}{2}[(a+b)^2+(b+c)^2+(a+c)^2]=a^2+b^2+c^2+ab+ac+bc\le 0$$
so we have
$$a=-b,b=-c,c=-a\Longrightarrow a=b=c=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Use integration by substitution I'm trying to evaluate integrals using substitution. I had
$$\int (x+1)(3x+1)^9 dx$$
My solution: Let $u=3x+1$ then $du/dx=3$
$$u=3x+1 \implies 3x=u-1 \implies x=\frac{1}{3}(u-1) \implies x+1=\frac{1}{3}(u+2) $$
Now I get $$\frac{1}{3} \int (x+1)(3x+1)^9 (3 \,dx) = \frac{1}{3} \int \frac{1}{3}(u+2)u^9 du = \frac{1}{9} \int (u+2)u^9 du \\
= \frac{1}{9} \int (u^{10}+2u^9)\,du = \frac{1}{9}\left( \frac{u^{11}}{11}+\frac{2u^{10}}{10} \right) + C$$
But then I get to this one
$$\int (x^2+2)(x-1)^7 dx$$
and the $x^2$ in brackets is throwing me off.
I put $u=x-1\implies x=u+1,$ hence $x^2+2 =(u+1)^2 +2 = u^2+3$. So
$$ \int(x^2+2)(x-1)^7\, dx = \int (u+1)u^7 du = \int (u+u^7) du = \frac{u^2}{2}+\frac{u^8}{8} + C $$
Is this correct or have I completely missed the point?
| Nice work, (though a wee bit hard to read).
Let's look a bit more closely at the second integral.
Your substitution is correct: $$u = x-1 \implies x = u+1,\;dx = du, \\ x^2 + 2 =(u+1)^2 + 2$$
You got a bit mixed up when substituting into the factor $x^2 + 2$, to express it as a function of $u$.
Substituting, we get $$\begin{align}\int (x^2 + 2)(x-1)^7\,dx & = \int ((u+1)^2 + 2)u^7\,du \\ \\
&= \int(u^2 + 2u + 3)u^7\,du \\ \\
&= \int(u^9 + 2u^8+ 3u^7)\,du\end{align}$$
I trust you can take it from here.
| {
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"url": "https://math.stackexchange.com/questions/1120516",
"timestamp": "2023-03-29T00:00:00",
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Let a,b,c be positive real numbers numbers such that $ a^2 + b^2 + c^2 = 3 $ Let $a,b,c\in\mathbb{R^+}$ such that $ a^2 + b^2 + c^2 = 3 $. Prove that
$$
(a+b+c)(a/b + b/c + c/a) \geq 9.
$$
My Attempt
I tried AM-GM on the symmetric expression so the $a+b+c \geq 3$, but I found $a+b+c \leq 3$.
| Consider the following optimization problem
$$\min_{x,y,z}\{(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\}$$
subject to
$$x^2+y^2+z^2=3$$
All we need to show is that
$$\min_{x,y,z}\{(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\}=9$$
under the condition
$$x^2+y^2+z^2=3$$
Set up the Lagrangian
$$\mathcal{L}(x,y,z,\lambda)=(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})+\lambda(3-x^2-y^2-z^2)$$
where $\lambda>0$.
First order conditions wrt $x,y,z$ yield
$$\mathcal{L}_x=0\Leftrightarrow(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})+(x+y+z)(\frac{1}{y}-\frac{z}{x^2})=2\lambda x$$
$$\mathcal{L}_y=0\Leftrightarrow(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})+(x+y+z)(\frac{1}{z}-\frac{x}{y^2})=2\lambda y$$
$$\mathcal{L}_z=0\Leftrightarrow(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})+(x+y+z)(\frac{1}{x}-\frac{y}{z^2})=2\lambda z$$
Exploiting the symmetry of the system of simultaneous equations above one obtains
$$x^*=y^*=z^*$$
where $(x^*,y^*,z^*)$ is the solution to the minimization problem. Invoking here the constraint we get
$$(x^*)^2+(y^*)^2+(z^*)^2=9\Rightarrow 3(x^*)^2=9\Rightarrow x^*=1$$
($x^*=1$,since $x^*\geq0$).
So $(x^*,y^*,z^*)=(1,1,1)$ minimizes the objective function i.e.
$$\min_{x,y,z}\{(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\}=(x^*+y^*+z^*)(\frac{x^*}{y^*}+\frac{y^*}{z^*}+\frac{z^*}{x^*})=(1+1+1)(\frac{1}{1}+\frac{1}{1}+\frac{1}{1})=9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How many 20-digit numbers are there which are formed using only the digits 5 and 7 and divisible by both 5 and 7. now i realised last digit has to be $5$ and position of $7$ wont affect divisibility.
$$x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}x_{7}x_{8}x_{9}x_{10}x_{11}x_{12}x_{13}x_{14}x_{15}x_{16}x_{17}x_{18}x_{19}x_{20}\equiv 0 \mod 7$$
so $x_{i}$ will be $1$ if it is $5$ and $0$ if it is $7$,so $x_{20}$=1.and then by remainder of powers of 10 i have to decide in which position $5$'s have to be placed.
so now its basically a combinatorics problem which i am unable to solve. please help.
| The question boils down to the following: How many $x_i \in \{ 5,7 \}$ can you find such that
$$x_1...x_{19}5 \equiv 0 \pmod{7}$$
Since $7$ is divisible by $7$, we can replace all $7's$ by $0$s.
Define $y_i =1$ if $x_i=5$ and $0$ otherwise. So the problem boils to
$$5\cdot y_1...y_{19}1 \equiv 0 \pmod{7}$$
or
$$ y_1...y_{19}1 \equiv 0 \pmod{7} \,.$$
As the powers of $10 pmod {7}$ are cyclically $1, 3, 2, 6, 4, 5$, the problem becomes
$$(1+y_{14}+y_{8}+y_{2})+3 (y_{19}+y_{13}+y_{7}+y_{1}) + 2 (y_{18}+y_{12}+y_{6})+6(y_{17}+y_{11}+y_{5})+4(y_{16}+y_{10}+y_{5})+5(y_{15}+y_{9}+y_{3}) \equiv 0 \pmod{7}$$
Define
$$a= 1+y_{14}+y_{8}+y_{2} \\
b=y_{19}+y_{13}+y_{7}+y_{1} \\
c=y_{18}+y_{12}+y_{6} \\
d= y_{17}+y_{11}+y_{5} \\
e=y_{16}+y_{10}+y_{5} \\
f= y_{15}+y_{9}+y_{3} (*)$$
Then the problem reduces to the following simpler problems:
Problem: Find all solutions of
$$a+3b+2c+6d+4e+5f =0 \pmod{7}$$
with $1 \leq a \leq 4 \,;\, 0 \leq b \leq 4 \,;\, 0 \leq c,d,e,f \leq 3$.
Now for each solution, when you look back at $(*)$ you have to figure out how many ways can you distribute the 1's among the y's. The problem becomes the problem of choosing in order $a-1, b, c, d, e, f$ positions for ones out of $3,4,3,3,3,3$, so the final answer will be
$$\sum_{(a,b,c,d,e,f) \mbox{solution} } \binom{3}{a-1} \binom{4}{b} \binom{3}{c}\binom{3}{d}\binom{3}{e}\binom{3}{f}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Weird integration issue: $\ln(x+1)=\ln(2x+2)$ ?! Weird integration issue: Using $(\ln[f(x)])'=\frac {f'(x)}{f(x)}$ we get that $\int \frac{2\,dx}{2x+2}=\ln(2x+2)$. Yet, $\int \frac{2\,dx}{2x+2}= \int\frac{dx}{x+1}=\ln(x+1)$ using the same rule as earlier.
What is wrong here?
| Remember that:
$$\int \frac{2}{2x+2}dx \neq \ln(2x+2)$$
But:
$$\int \frac{2}{2x+2}dx = \ln(2x+2)+C$$
where $C$ is constant, so:
$$\int \frac{2}{2x+2}dx = \ln(2x+2)+C=\ln(2(x+1))+C=\ln(x+1)+\ln 2+C=\ln(x+1)+(\ln 2+C)=\ln(x+1)+C_1$$
So both calculations are almost correct, but you forgot about constant.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of $\lfloor k^{1/3} \rfloor$ I am faced with the following sum:
$$\sum_{k=0}^m \lfloor k^{1/3} \rfloor$$
Where $m$ is a positive integer. I have determined a formula for the last couple of terms such that $\lfloor n^{1/3} \rfloor^3 = \lfloor m^{1/3} \rfloor^3$. For example if the sum is from 0 to 11 I can find the sum of the terms 8 through 11. I am stuck on what formula can be applied to find the sum of the terms before the final stretch however.
| You have
$$
\begin{align*}
\sum_{k=0}^m \lfloor k^{1/3} \rfloor &= \sum_{i=0}^{\lfloor m^{1/3} \rfloor} i |\{0 \leq k \leq m : \lfloor k^{1/3} \rfloor = i\}| \\ &=
\sum_{i=0}^{\lfloor m^{1/3} \rfloor} i |\{0 \leq k \leq m : i^3 \leq k < (i+1)^3\}| \\ &=
\sum_{i=0}^{\lfloor m^{1/3} \rfloor-1} i((i+1)^3-i^3) + \lfloor m^{1/3} \rfloor (m-\lfloor m^{1/3} \rfloor^3+1) \\ &=
\sum_{i=0}^{\lfloor m^{1/3} \rfloor-1} (3i^3+3i^2+i) + \lfloor m^{1/3} \rfloor (m-\lfloor m^{1/3} \rfloor^3+1) \\ &=
\frac{1}{4}(\lfloor m^{1/3} \rfloor-1)\lfloor m^{1/3} \rfloor^2(3\lfloor m^{1/3} \rfloor+1) + \lfloor m^{1/3} \rfloor (m-\lfloor m^{1/3} \rfloor^3+1).
\end{align*}
$$
| {
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Determine the value of k for which a matrix system is consistent and the values for which it is inconsistent The non-homogenous system is as follows:
$$3x+2y+5z=10\\
3x-2y=7\\
6x+4y-10z=k$$
I have determined that:
$$z=1-\frac{k}{20}\\
y=\frac{k}{16}-\frac{1}{2}\\
x=2+\frac{k}{24}$$
What are the values of $k$ to form a matrix which is inconsistent and then consistent?
| Let $K$ be a field and $k\in K$. Then we can write the system of linear equations in matrix form $Av=b$ with
$$
A=\begin{pmatrix} 3 & 2 & 5 \\ 3 & -2 & 0 \\ 6 & 4 & -10 \end{pmatrix},\quad v=\begin{pmatrix}
x \\ y \\ z \end{pmatrix},\quad b=\begin{pmatrix}
10 \\ 7 \\ k \end{pmatrix}
$$
The system has a unique solution iff $\det(A)\neq 0$ in $K$. In this case the solution is $v=A^{-1}b$. Since $\det(A)=120$ is nonzero for all fields of characteristic not equal to $2,3,5$, we obtain a unique solution
$$
v=A^{-1}b=\begin{pmatrix} \frac{48+k}{24} \\ \frac{k-8}{16} \\ \frac{20-k}{20} \end{pmatrix}
$$
for any given $k\in K$. The system may be inconsistent for characteristic $2,3,5$. For example, the last equation for $k=1$ and $char(K)=2$ is inconsistent: $0=1$.
| {
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Proof: $n^p < \frac{(n+1)^{p+1}-n^{p+1}}{p+1} < (n+1)^p$ I've edited the post in order to add at the end what, I think, is the complete proof of these inequalities. I want to apologize by not having given a reply as early as those given by the users who gave me hints about this exercise. It took me a little of time but your answers were very helpful (:.
This is actually part (b) of the exercise. I was able to prove these inequalities by applying the Binomial Theorem, but I have no idea about how to do it using part (a), which I'll post here too.
(a) Let $b$ be a positive integer. Prove that:
$$b^p - a^p = (b-a)(b^{p-1}+ b^{p-2}a+b^{p-3}a^{2}+...+ba^{p-2}+a^{p-1})$$
(b) Let $p$ and $n$ denote positive integers. Use part (a) to show that
$$n^p < \frac{(n+1)^{p+1}-n^{p+1}}{p+1} < (n+1)^p$$
(a) Proof:
$$\begin{split}b^p - a^p &= (b-a)(b^{p-1}+ b^{p-2}a+b^{p-3}a^{2}+...+ba^{p-2}+a^{p-1})\\
&= (b-a)\sum_{k=0}^{p-1}\big[b^{p-(k+1)}a^k\big]\\
&=\sum_{k=0}^{p-1}\big[b^{p-(k+1)+1}a^k- b^{p-(k+1)}a^{k+1}\big]\quad \Leftarrow \small{\text{Distribute }(b-a).}\\
&=\sum_{k=0}^{p-1}\big[b^{p-k}a^k- b^{p-(k+1)}a^{k+1}\big]\\
&=b^{p-0}a^0- b^{p-[(p-1)+1]}a^{(p-1)+1}\quad\Leftarrow \small{\text{Apply the telescoping property for sums.}}\\
&= b^p - a^p\end{split}$$
(b) This is my attempt:
$$n^p < \frac{(n+1)^{p+1}-n^{p+1}}{p+1} < (n+1)^p$$
By multiplying by $p+1$ we have
$$(p+1)n^p < (n+1)^{p+1}-n^{p+1} < (p+1)(n+1)^p$$
which can also be written as
$$(p+1)n^p < [(n+1)-n]\sum_{k=0}^{p}(n+1)^{p-(k+1)}n^{k} < (p+1)(n+1)^p\\
(p+1)n^p < \sum_{k=0}^{p}(n+1)^{p-(k+1)}n^{k} < (p+1)(n+1)^p$$
By dividing the inequalities by $(n+1)^p$ we get
$$(p+1)\left(\frac{n}{n+1}\right)^p < \sum_{k=0}^{p}\frac{n^k}{(n+1)^{k+1}} < (p+1)$$
Here I'm stuck. I guess the last step wasn't necessary.
Edit:
This is my last attempt. Hopefully it's not flawed.
We will prove each bound separately.
To prove: $(p+1)n^p < (n+1)^{p+1}-n^{p+1}$
Proof (direct):
Let the numbers $a$ and $b$ be defined as
$$a = n \quad \text{and}\quad b = n+1\qquad \text{For }n \in \mathbb{N}.$$
Then we have
$$a < b$$
And it follows
$$\frac{a^p}{a^k} < \frac{b^p}{b^k}\quad \text{For } p\ \text{and }k\in\mathbb{N}, p \neq k. \qquad (1)$$
By multiplying both sides by $a^k$ we get
$$a^p < b^{p-k}a^k$$
By taking the sum of both sides we have
$$\sum_{k=0}^pa^p < \sum_{k=0}^pb^{p-k}a^k$$
On the LHS $a^p$ is summed $p+1$ times (from $0$ to $p$). So it can also be written as
$$(p+1)a^p < \sum_{k=0}^pb^{p-k}a^k$$
Since $b-a = 1$, let multiply the RHS by $b-a$
$$(p+1)a^p < (b-a)\sum_{k=0}^pb^{p-k}a^k$$
By distributing $b-a$ inside the sum we have
$$(p+1)a^p <\sum_{k=0}^p[b^{p-(k-1)}a^k - b^{p-k}a^{k+1}]$$
And by applying the telescoping property for sums we get
$$\begin{align*}(p+1)a^p &< b^{p-(0-1)}a^0 - b^{p-p}a^{p+1}\\
(p+1)a^p &< b^{p+1} - a^{p+1}\\
(p+1)n^p &< (n+1)^{p+1} - n^{p+1}\end{align*}$$
which completes the proof.
To prove: $(n+1)^{p+1}-n^{p+1} < (n+1)^p$
Proof (direct):
By the definition of $a$ and $b$ previously given and from inequality $(1)$ we have
$$\frac{a^p}{a^k} < \frac{b^p}{b^k}$$
Let multiplying both sides by $b^k$
$$a^{p-k}b^k < b^p$$
Let take the sum of both sides
$$\sum_{k=0}^pa^{p-k}b^k < \sum_{k=0}^pb^p$$
On the RHS $b^p$ is summed $p+1$ times. Then
$$\sum_{k=0}^pa^{p-k}b^k < (p+1)b^p$$
Let multiply the LHS by $1 = b-a$
$$\begin{align*}(b-a)\sum_{k=0}^pa^{p-k}b^k < (p+1)b^p\\
\sum_{k=0}^p[a^{p-k}b^{k+1}-a^{p-(k-1)}b^k] < (p+1)b^p\end{align*}$$
And by applying the telescoping property for sums we get
$$\begin{align*}a^{p-p}b^{p+1}-a^{p-(0-1)}b^0 &< (p+1)b^p\\
b^{p+1}-a^{p+1} &< (p+1)b^p\\
(n+1)^{p+1}-n^{p+1} &< (p+1)(n+1)^p\end{align*}$$
which completes the proof.
It looks like I can go to sleep without remorse, doesn't it? (:
| The inequality $\dfrac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^{p+1}$ is trivial. You obtain the left-hand-side by subtracting from the right and dividing by a number larger than $1$.
On the other hand, if you use part (a) you have
$$(n+1)^{p+1} - n^{p+1} = (1) \bigg((n+1)^p + (n+1)^{p-1}n + \cdots + (n+1)n^{p-1} + n^p\bigg).$$
Each term on the right is $\ge n^p$, and there are $p+1$ of them. Thus $$(n+1)^{p+1} - n^{p+1} >(p+1)n^p.$$
| {
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"url": "https://math.stackexchange.com/questions/1143437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Integral $\int\frac{dx}{x^5+1}$ Calculate $\displaystyle\int\dfrac{dx}{x^5+1}$
| One might as well explain how to derive the factorization in Troy Woo's answer. (This is by hand, not software.) Yes, he's right that any polynomial with real coefficients can be factorized into linear and quadratic factors with real coefficients; this fact can be derived from the fundamental theorem of algebra, by factorizing into linear factors $x - r_i$ where the $r_i$ range over all complex roots, and collecting factors that involve non-real roots into complex conjugate pairs in order to get at the real quadratic factors.
Applying this to the polynomial $x^5 + 1$, let $\zeta$ be a complex fifth root of $-1$, say $\zeta = e^{\pi i/5}$. The other complex (non-real) fifth roots are $\bar{\zeta} = \zeta^{-1}$, $\zeta^3 = -\zeta^{-2}$, and $\bar{\zeta^3} = -\zeta^2$. Then
$$x^5 + 1 = (x+1)(x - \zeta)(x - \zeta^{-1})(x + \zeta^2)(x + \zeta^{-2})$$
where we may observe
$$0 = 1 - \zeta - \zeta^{-1} + \zeta^2 + \zeta^{-2}$$
by matching coefficients of $x^4$ on both sides. One of the conjugate pairs gives a real quadratic factor $(x - \zeta)(x - \zeta^{-1}) = x^2 - \alpha x + 1$. Here the real number $\alpha = \zeta + \zeta^{-1}$ can be computed as follows:
$$\alpha^2 = (\zeta + \zeta^{-1})^2 = \zeta^2 + 2 + \zeta^{-2} = 2 + (-1 + \zeta + \zeta^{-1}) = 1 + \alpha$$
where the third equation is by the observation above. Solving for $\alpha$ gives either $\frac{1 + \sqrt{5}}{2}$ or $\frac{1 - \sqrt{5}}{2}$; actually it's the first if we start with the fifth root $\zeta = e^{\pi i/5}$. The other conjugate pair leads to $(x + \zeta^2)(x + \zeta^{-2}) = x^2 - \beta x + 1$ where $\beta$ is the other root $\frac{1 - \sqrt{5}}{2}$. This gives Troy Woo's answer.
| {
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Proving $\sum_{n=0}^{\infty }\frac{\sin^4(4n+2)}{(2n+1)^2}=\frac{5\pi ^2}{16}-\frac{3\pi }{4}$ I dont have an idea to prove it because of exist $\sin(4n+2)^4$
$$\sum_{n=0}^{\infty }\frac{\sin^4(4n+2)}{(2n+1)^2}=\frac{5\pi ^2}{16}-\frac{3\pi }{4}$$
| Since:
$$\sin x = \frac{1}{2i}\left(e^{ix}-e^{-ix}\right)\tag{1}$$
we have:
$$\sin^4 x = \frac{1}{16}\left(e^{4ix}+e^{-4ix}-4e^{2ix}-4e^{2ix}+6\right)=\frac{1}{8}\left(\cos(4x)-4\cos(2x)+3\right).\tag{2} $$
It follows that:
$$\begin{eqnarray*} S &=& \sum_{n\geq 0}\frac{\sin^4(4n+2)}{(2n+1)^2}\\&=&\frac{1}{8}\sum_{n\geq 0}\frac{\cos(16n+8)}{(2n+1)^2}-\frac{1}{2}\sum_{n\geq 0}\frac{\cos(8n+4)}{(2n+1)^2}+\frac{3}{8}\sum_{n\geq 0}\frac{1}{(2n+1)^2}\tag{3}\end{eqnarray*}$$
where $\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$. Now it happens that:
$$ f(x) = \sum_{n\geq 0}\frac{\cos((2n+1)x)}{(2n+1)^2}\tag{4}$$
is the Fourier series of the triangle wave with period $2\pi$ and amplitude $2\cdot \frac{\pi^2}{8}$, hence:
$$ f(4)=\pi-3\frac{\pi^2}{8}, \qquad f(2)=-\frac{\pi}{2}+\frac{\pi^2}{8} \tag{5}$$
and:
$$ S = \frac{1}{8}f(4)-\frac{1}{2}f(2)+\frac{3}{8}f(0)=\color{red}{\frac{5\pi^2}{16}-\frac{3\pi}{4}}\tag{6}$$
as wanted.
Update. Once proved that $\sin^4(2x)=\frac{1}{2}\left(1-\cos(2x)\right)-\frac{1}{8}\left(1-\cos(4x)\right),$ we have:
$$ S = \sum_{n\geq 0}\frac{\sin^4(2n)}{n^2}\cdot\frac{1-\cos(\pi n)}{2}=\sum_{n\geq 0}\frac{\sin^2(n)\sin^2\left(\frac{\pi}{2}n\right)}{n^2}-\frac{1}{4}\sum_{n\geq 0}\frac{\sin^2(2n)\sin^2\left(\frac{\pi}{2}n\right)}{n^2} $$
and the last sums can be computed through Parseval's theorem.
From the Gudermannian function we know that:
$$\sum_{n\geq 0}\sin\left(\frac{\pi}{2}n\right)\frac{\sin(nx)}{n}=\frac{1}{2}\operatorname{arctanh}(\sin x)=\frac{1}{4}\log\left(\frac{1+\sin x}{1-\sin x}\right) $$
and we are left with:
$$\frac{2}{\pi}\int_{0}^{\pi}\left(\frac{1}{4}\log\left(\frac{1+\sin x}{1-\sin x}\right)\right)^2\,dx =\frac{2}{\pi}\int_{0}^{1}\log^2\left(\frac{1+t}{1-t}\right)\frac{dt}{1+t^2}=\frac{2}{\pi}\int_{0}^{+\infty}\frac{\log^2 z}{1+z^2}\,dz$$
where the last integral equals $\frac{\pi^3}{8}$ by the residue theorem.The other integral is even easier.
| {
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"source": "stackexchange",
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$\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$, where numbers $a$ and $b$ are rational If $a$ and $b$ are rational numbers such that $\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$
Then what is the value of $a$? The answer is $-1$.
$$\sqrt{4 - 2\sqrt{3}} = a + b\sqrt{3}$$
$$4 - 2\sqrt{3} = 2^2 - 2\sqrt{3}$$ Let $u =2$ hence,
$$\sqrt{u^2 - \sqrt{3}u} = a + b\sqrt{3}$$
$$u^2 - \sqrt{3}u = u(u - \sqrt{3})$$
$$a + b\sqrt{3} = \sqrt{u}\sqrt{u - \sqrt{3}}$$
What should I do?
| Such square roots can be computed by a Simple Denesting Rule:
Here $\ 4-2\sqrt 3\ $ has norm $= 4.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = 2\,\ $ yields $\,\ 2-2\sqrt 3\:$
which has $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{4}\, =\, 2.\ \ \rm \color{brown}{Dividing\ it\ out}\ $ of the above yields $\ \ 1-\sqrt 3$
Finally, since the result is negative, we need to negate it, which yields $\ \sqrt 3 - 1$
Remark $\ $ Many more worked examples are in prior posts on this denesting rule.
| {
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Exponential Generating Functions (EGF) (a) Find the exponential generating function (in closed form) for the number of ways to place n students into 3 classes (one MATH, one BIOLOGY and one PHYSICS), such that the MATH class does not have exactly one student. (Classes are allowed to be empty).
(b) Using the e.g.f. found in part (a). Find the formula for the coefficient of $\frac{x^n}{n!}$.
Workings:
$f(x) = (1 + \frac{x^2}{2!} + \frac{x^3}{3!} + ...) (1 + x + \frac{x^2}{2!} + ...) (1 + x + \frac{x^2}{2!} + ...)$
$f(x) = (e^x - x)e^xe^x$
$f(x) = (e^x-x)e^{2x}$
b.
$f(x) = (e^x-x)e^{2x}$
$f(x) = e^{3x}-xe^{2x}$
$e^{3x}-xe^{2x} = \sum\limits_{n=0}^{\infty}3^n\frac{x^n}{n!}-x\sum\limits_{r=0}^{\infty}2^n\frac{x^n}{n!}$
The coefficient of $\frac{x^n}{n!}$ is $3^n - x*2^n$.
I'm wondering if I did this correctly.
Any help will be appreciated.
| You were doing fine until you tried to extract the coefficient of $\frac{x^n}{n!}$ from $xe^{2x}$. The coefficient is just that: it cannot involve $x$. Thus, you need to incorporate that extra factor of $x$ in the summation:
$$x\sum_{n\ge 0}2^n\frac{x^n}{n!}=\sum_{n\ge 0}2^n\frac{x^{n+1}}{n!}\;.$$
The problem now is that we have $\frac{x^{n+1}}{n!}$ and not $\frac{x^n}{n!}$. As a first step, notice that
$$\frac{x^{n+1}}{n!}=(n+1)\frac{x^{n+1}}{(n+1)!}\;,$$
so we can rewrite the summation as
$$\sum_{n\ge 0}2^n(n+1)\frac{x^{n+1}}{(n+1)!}$$
and do a simple index shift to get
$$\sum_{n\ge 1}n2^{n-1}\frac{x^n}{n!}\;.$$
And since $n2^{n-1}=0$ when $n=0$, we can start the summation at $n=0$ without doing any harm. Thus,
$$e^{3x}-xe^{2x}=\sum_{n\ge 0}3^n\frac{x^n}{n!}-\sum_{n\ge 0}n2^{n-1}\frac{x^n}{n!}=\sum_{n\ge 0}\left(3^n-n2^{n-1}\right)\frac{x^n}{n!}\;,$$
and the coefficient of $\frac{x^n}{n!}$ is $3^n-n2^{n-1}$.
Added: Note that you can do some checking of your answer. By brute force calculation we have
$$\begin{align*}
f(x)&=\left(1+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\right)e^{2x}\\
&=\left(1+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\right)\left(1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}+\ldots\right)\\
&=1+2x+5\frac{x^2}{2!}+15\frac{x^3}{3!}+49\frac{x^4}{4!}+\ldots\;,
\end{align*}$$
and the coefficients $1,2,5,15$, and $49$ are indeed those given by the formula.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$
A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\geq3\prod_{cyc}(a+b-c),$$ or
$$\sum_{cyc}(a^3-ab^2-ac^2+2abc)\geq\sum_{cyc}(-3a^3+3a^2b+3a^2c-2abc),$$ but it becomes very ugly.
| Suppose that $S>0$. Then for $x\in(0,S/2)$, the function $f(x)=\frac{x}{S-2x}$ is convex. Thus, by Jensen's inequality and with $S=a+b+c$, we have
$$
\frac{1}{3}f(a)+\frac{1}{3}f(b)+\frac{1}{3}f(c)\geq f\left(\frac{1}{3}(a+b+c)\right)=\frac{S/3}{S-2S/3}=1.
$$
This is equivalent to $f(a)+f(b)+f(c)\geq 3$, which is your inequality.
| {
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How to calculate derivative of $f(x) = \frac{1}{1-2\cos^2x}$? $$f(x) = \frac{1}{1-2\cos^2x}$$
The result of $f'(x)$ should be equals
$$f'(x) = \frac{-4\cos x\sin x}{(1-2\cos^2x)^2}$$
I'm trying to do it in this way but my result is wrong.
$$f'(x) = \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} =
\frac {1-2\cos^2x-(1-(2\cos^2x)')}{(1-2\cos^2x)^2} = $$
$$=\frac {-2\cos^2x + 2(2\cos x(\cos x)')}{(1-2\cos^2x)^2} =
\frac {-2\cos^2x+2(-2\sin x\cos x)}{(1-2\cos^2x)^2} = $$
$$\frac {-2\cos^2x-4\sin x\cos x}{(1-2\cos^2x)^2}$$
| $f(x)=\dfrac{1}{1-2\cos^2 x}=-\dfrac{1}{2\cos^2 x-1}=-\dfrac{1}{\cos 2x}=-\sec 2x$
$\therefore f'(x)=-2\sec(2x)\tan(2x) \quad\blacksquare$
Check the list of formulas for finding derivatives of trigonometric functions here
| {
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Discrete-time sinusoids with same frequency I've read that sine waves of the form $x_n = \sin(w_{0}n)$, with frequencies $w_{0}$ and $w_{0} + 2\pi$, are indistinguishable from each other when considering discrete time.
The book gives $\cos\big(\frac{\pi}{4}n\big)$ as an example. Adding $2\pi$ to the frequency, we get $\cos\big(\frac{9\pi}{4}n\big)$ which is indeed the same wave. However, the text's answer is $\cos\big(\frac{7\pi}{4}n\big)$ which is again the same wave I plot below. Which calculations lead to this answer?
Thanks in advance!
| For $n$ an integer,
$$\cos\frac{7\pi n}{4}= \cos\frac{9\pi n}{4} $$
because for $n$ even both are simply zero, and for $n$ odd, $\frac{7\pi n}{4}$ is $-\left( \frac{9\pi n}{4} \right) + 2\pi$ but
$$
\cos\left[-\left( \frac{9\pi n}{4} \right) + 2\pi \right]= \cos[-\left( \frac{9\pi n}{4} \right)] = \cos\left[+\left( \frac{9\pi n}{4} \right)\right] $$
because cosine is an even function of its argument.
So both answers give the same graph.
| {
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An $\varepsilon-\delta$ proof of $\lim_{x\rightarrow c} (2x^ 2 − 3x + 4) = 2c ^2 − 3c + 4$.
Give an $\varepsilon-\delta$ proof using the “$\delta\leq 1 $ trick” to prove that $\lim_{x\rightarrow c} (2x^ 2 − 3x + 4) = 2c ^2 − 3c + 4$.
My attempt: Let $ε>0$ and there exist $\delta>0$ such that whenever $|x-c|<δ$,
$$|2x^ 2 − 3x + 4-(2c ^2 − 3c + 4)|<ε$$
$$\implies |2x^2-2c^2-3x+3c|<ε$$
$$\implies 2|x^2-c^2|-3|x-c|<ε$$
$$\implies 2|x-c||x+c|-3|x-c|<ε$$
$$\implies |x-c||2x+2c-3|<ε$$
I don't see where to go from here.
| Note that
$$|x+c|\leq |x-c|+|2c|$$
which means
$$2|x-c||x+c|+3|x-c|\leq 2|x-c|^2+4|c||x-c|+3|x-c|=|x-c|(2|x-c|+4|c|+3)$$
Choose
$$\delta:=\min\left\{\frac{\epsilon}{8|c|+6}, 2|c|+\frac{3}{2}\right\}$$
Then for $|x-c|<\delta$, we have
$$|x-c|(2|x-c|+4|c|+3)<\delta(2\delta+4|c|+3)\leq\frac{\varepsilon}{8|c|+6}\left(2\left(2|c|+\frac{3}{2}\right)+4|c|+3\right)=\epsilon$$
which gives
$$|(2x^2-3x+4)-(2c^2-3c+4)|\leq |x-c|(2|x-c|+4|c|+3)<\epsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1162785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplify $\frac{\cos \theta}{\sec \theta} + \frac{\sin \theta}{\csc \theta}$ Please show how to simplify this type of expression without using a calculator. I'm new to trigonometry and I don't know how to simplify this expressions: $\frac{\cos \theta}{\sec \theta} + \frac{\sin \theta}{\csc \theta}$.
| First, we need to make a common denominator of $\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta}{1+\cos\theta}$.
Let's go!
$$\require{cancel}\begin{aligned}\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta}{1+\cos\theta}&=\frac{\sin\theta(1+\cos\theta)-\sin\theta(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}\\&=\frac{\cancel{\sin\theta}+\sin\theta\cos\theta\cancel{-\sin\theta}+\sin\theta\cos\theta}{1-\cos^2\theta}\\&=\frac{2\cancelto{1}{\sin\theta}\cos\theta}{\cancelto{\sin\theta}{\sin^2\theta}}\\&=\frac{2\cos\theta}{\sin\theta}\\&=2\cot\theta\end{aligned}$$
I hope this helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trick with modular exponentiation For example,
$$123^{25} \pmod{10}$$
$$ 123 \equiv 3 \pmod{10}$$
$$123^{2} \equiv 9 \pmod{10}$$
$$123^{3} \equiv 7 \pmod{10}$$
$$123^{4} \equiv 1 \pmod{10}$$
$$123^{5} \equiv 3 \pmod{10}$$
It is easy to see:
$$123^n \equiv 123^{4(k) + n} \pmod{10}$$
$$4(6) = 4(k) = 24$$ Hence,
$$123^{1} \equiv 123^{4(6) + 1} \equiv 123^{25} \pmod{10} \equiv 3 \pmod{10}$$
My question is why does this work?
| First of all $123^n = (120+3)^n = (12\times10+3)^n = \sum_1^n \binom{n}{k}\times(12\times10)^k\times3^{n-k} \equiv 3^n \pmod {10}$, because all multiples of $10 \equiv 0 \pmod {10}$. So it's only needed to solve $3^{25} \pmod {10}$
This reduces it to $3^n \equiv 3^{n+4k} \pmod{10}$. Why $4$? As @André Nicolas stated, this is because of Euler's theorem saying $a^{\varphi(m)}\equiv 1 \pmod m$, if $\mathrm{gcd}(a,m) = 1$. This means $3^n \pmod {10}$ will only hit numbers coprime to $10$, and these are the four numbers $\{1,\,3,\,7,\,9\}$. None of these numbers share a divisor with $10$. This also means that $3^n \pmod {10}$ cannot hit $\{0,\,2,\,4,\,5,\,6,\,8\}$, since for instance there is no number $n$ such that $3^n = 4 \pmod{10}$. So since it can only hit the four numbers $\{1,\,3,\,7,\,9\}$, it will only do so, and repeat once it has hit the four numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that the limit of the following function of two variables is zero I need to prove the following:
$$\lim_{(x,y)\to (1 ,2)} \frac{x^2+2xy-6x-2y+5}{\sqrt{(x-1)^2+(y-2)^2}}=0$$
I've tried to solve it by substituting $y=mx$ but I can't get the solution that way. Any help would be appreciated thanks.
| Factor the numerator as $$x^2+2xy-6x-2y+5=(x-1)[(x-1)+2(y-2)]$$Then, use the triangle inequality$$|(x-1)[(x-1)+2(y-2)]|\le|(x-1)^2|+2|x-1||y-2|$$to show that$$\frac{|x^2+2xy-6x-2y+5|}{\sqrt{(x-1)^2+(y-2)^2}}\le\frac{|(x-1)^2|+2|x-1||y-2|}{\sqrt{(x-1)^2+(y-2)^2}}$$Noting that $\sqrt{(x-1)^2+(y-2)^2}\ge|x-1|$ $$\frac{|x^2+2xy-6x-2y+5|}{\sqrt{(x-1)^2+(y-2)^2}}\le\frac{|(x-1)^2|+2|x-1||y-2|}{\sqrt{(x-1)^2+(y-2)^2}}\le|x-1|+2|y-2|$$
Finally, for all $\epsilon>0$ $$\frac{|x^2+2xy-6x-2y+5|}{\sqrt{(x-1)^2+(y-2)^2}}<\epsilon$$whenever $|x-1|<\delta_1=\frac{\epsilon}{3}$ and $|y-2|<\delta_2=\frac{\epsilon}{3}$.
| {
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Find $\lim \limits_{(x,y) \rightarrow (0,0)}\frac{x^3 y^3 }{x^2+y^2}$ (epsilon delta proof) Suppose we have the following function:
$f(x,y)=\dfrac{x^3 y^3 }{x^2+y^2}$.
Determine $\lim \limits_{(x,y) \rightarrow (0,0)}f(x,y)$.
I did the following, but I cannot continue.
Suppose $0 < \sqrt{x^2+y^2} < \delta$.
$$\left|\frac{x^3 y^3}{x^2 + y^2}-0\right| = \left|\frac{xyx^2y^2}{x^2+y^2}\right| \le \left|\frac{(x^2+y^2)^2(xy)}{x^2+y^2}\right|=|xy|\left|x^2+y^2\right|<|xy|\delta^2=\epsilon.$$ The $|xy|$ makes it impossible for me to relate $\delta$ to $\epsilon$. How can I solve it?
| Hint: $\forall (x,y)\in \mathbb R^2\left(2|xy|\leq x^2+y^2\right)$.
This inequality helps not only for finding $\delta$, but also to get $$\forall (x,y)\in \mathbb R^2\left((x,y)\neq (0,0)\implies \left|\dfrac{xyx^2y^2}{x^2+y^2}\right|= \dfrac{x^2y^2}2\left|\dfrac{2xy}{x^2+y^2}\right|\leq \dfrac{x^2y^2}{2}\right),$$
which deviates slightly from your work.
| {
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Proof of an Limit Using the formal definition of convergence, Prove that $\lim\limits_{n \to \infty} \frac{3n^2+5n}{4n^2 +2} = \frac{3}{4}$.
Workings:
If $n$ is large enough, $3n^2 + 5n$ behaves like $3n^2$
If $n$ is large enough $4n^2 + 2$ behaves like $4n^2$
More formally we can find $a,b$ such that $\frac{3n^2+5n}{4n^2 +2} \leq \frac{a}{b} \frac{3n^2}{4n^2}$
For $n\geq 2$ we have $3n^2 + 5n /leq 3n^2.
For $n \geq 0$ we have $4n^2 + 2 \geq \frac{1}{2}4n^2$
So for $ n \geq \max\{0,2\} = 2$ we have:
$\frac{3n^2+5n}{4n^2 +2} \leq \frac{2 \dot 3n^2}{\frac{1}{2}4n^2} = \frac{3}{4}$
To make $\frac{3}{4}$ less than $\epsilon$:
$\frac{3}{4} < \epsilon$, $\frac{3}{\epsilon} < 4$
Take $N = \frac{3}{\epsilon}$
Proof:
Suppose that $\epsilon > 0$
Let $N = \max\{2,\frac{3}{\epsilon}\}$
For any $n \geq N$, we have that $n > \frac{3}{\epsilon}$ and $n>2$, therefore
$3n^2 + 5n^2 \leq 6n^2$ and $4n^2 + 2 \geq 2n^2$
Then for any $n \geq N$ we have
$|s_n - L| = \left|\frac{3n^2 + 5n}{4n^2 + 2} - \frac{3}{4}\right|$
$ = \frac{3n^2 + 5n}{4n^2 + 2} - \frac{3}{4}$
$ = \frac{10n-3}{8n^2+4}$
Now I'm not sure on what to do. Any help will be appreciated.
| Note
$$\left|\frac{3n^2 + 5n}{4n^2 + 2} - \frac{3}{4}\right| = \left|\frac{20n - 6}{4(4n^2 + 2)}\right| = \frac{|10n - 3|}{4(2n^2 + 1)} < \frac{10n + 3}{8n^2} < \frac{20n}{8n^2} = \frac{5}{2n}.\tag{1}$$
Hence, given $\epsilon > 0$, setting $N > 2\epsilon/5$ will make the left hand side of $(1)$ less than $\epsilon$ whenever $n \ge N$. Therefore
$$\lim_{n\to \infty} \frac{3n^2 + 5n}{4n^2 + 2} = \frac{3}{4}.$$
| {
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Prove that $ \cos x - \cos y = -2 \sin ( \frac{x-y}{2} ) \sin ( \frac{x+y}{2} ) $ Prove that $ \cos x - \cos y = -2 \sin \left( \frac{x-y}{2} \right) \sin \left( \frac{x+y}{2} \right) $ without knowing cos identity
We don't know that $ \cos0 = 1 $
We don't know that $ \cos^2 x + \sin^2 x = 1 $
I have managed to prove it using the above facts, but just realised that I can't use them. Now I have been going in circles for a while.
Any ideas how to prove this or even approach it?
Thanks !
| Remember this trick for life: write $x = a+b$ and $y = a-b$. So: $$\begin{cases} x&= a+b \\ y &= a-b \end{cases} \implies \begin{cases} a &= \frac{x+y}{2}\\ b &= \frac{x-y}{2} \end{cases}.$$
Then: $$\begin{align} \cos x - \cos y &= \cos(a+b)-\cos(a-b) \\ &= \cos a \cos b - \sin a \sin b - (\cos a \cos b + \sin a \sin b) \\ &= -2 \sin a \sin b \\ &= -2 \sin \left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2} \right).\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral of square root. We have
$$\int \sqrt{x} \text{ }dx=\int x^{1/2} \text{ }dx=\frac{2}{3}x^{3/2}+C.$$
But, let $u=\sqrt{x}$. Then $x=u^2$, and $dx=2udu$. Substituting, we have
$$\int \sqrt{x} \text{ }dx=\int 2u^2 \text{ }du=\frac{2}{3}x^3+C.$$
Which one is correct, and why?
| You have $u= \sqrt{x}$, hence $\int 2u^2 du = \frac{2u^3}{3}+C= \frac{2x^{3/2}}{3}+C$
| {
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Prove the following conditional divisibility If $gcd(a,b)=1$ and $n$ is a prime number,then prove that $\frac{(a^n + b^n)}{(a+b)}$ and $(a+b)$ have no factors in common unless $(a+b)$ is a multiple of $n$.
I don't know how to establish the relation between $n$ and $(a+b)$. This is how much I have been able to derive :
$${\frac{(a^n+b^n)}{(a+b)}=(a+b)^{n-1} + \frac{C_1a^{n-1}b +C_2a^{n-2}b^2+...+C_{n-1}ab^{n-1}}{(a+b)}}$$
We need to prove that,$${gcd((a+b)^{n-1}+\frac{C_1a^{n-1}b +C_2a^{n-2}b^2+...+C_{n-1}ab^{n-1}}{(a+b)},(a+b))=1}$$ unless $n=s.(a+b)$ for some $s$.
I have no idea how to proceed. Please help. Thank you! :)
| Because $n$ is odd (note: for $n=2$, $\frac{a^2+b^2}{a+b}$ need not be an integer) we have $$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-a^{n-4}b^3+\cdots-a^1b^{n-2}+b^{n-1})$$
We compute $$gcd(a^{n-1}-a^{n-2}b+a^{n-3}b^2-a^{n-4}b^3+\cdots-a^1b^{n-2}+b^{n-1},a+b)=$$
$$gcd(a^{n-1}-a^{n-2}b+a^{n-3}b^2-a^{n-4}b^3+\cdots-a^1b^{n-2}+b^{n-1}\color{red}{-a^{n-2}(a+b)},a+b)=$$
$$gcd(-2a^{n-2}b+a^{n-3}b^2-a^{n-4}b^3+\cdots-a^1b^{n-2}+b^{n-1}\color{red}{+2a^{n-3}(a+b)},a+b)=$$
$$gcd(3a^{n-3}b^2-a^{n-4}b^3+\cdots-a^1b^{n-2}+b^{n-1}\color{red}{-3a^{n-4}(a+b)},a+b)=$$
$$\cdots$$
$$=gcd(nb^{n-1},a+b)$$
However, since $gcd(a,b)=1$, also $1=gcd(a+b,b)=gcd(a+b,b^{n-1})$. Hence we can reduce the original to $gcd(n,a+b)$. Since $n$ is prime this is either $1$ or $n$.
| {
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Proof that every number ≥ $8$ can be represented by a sum of fives and threes. Can you check if my proof is right?
Theorem. $\forall x\geq8, x$ can be represented by $5a + 3b$ where $a,b \in \mathbb{N}$.
Base case(s): $x=8 = 3\cdot1 + 5\cdot1 \quad \checkmark\\
x=9 = 3\cdot3 + 5\cdot0 \quad \checkmark\\
x=10 = 3\cdot0 + 5\cdot2 \quad \checkmark$
Inductive step:
$n \in \mathbb{N}\\a_1 = 8, a_n = a_1 + (x-1)\cdot3\\
b_1 = 9, b_n = b_1 + (x-1)\cdot3 = a_1 +1 + (x-1) \cdot 3\\
c_1 = 10, c_n = c_1 + (x-1)\cdot3 = b_1 + 1 + (x-1) \cdot 3 = a_1 + 2 + (x-1) \cdot 3\\
\\
S = \{x\in\mathbb{N}: x \in a_{x} \lor x \in b_{x} \lor x \in c_{x}\}$
Basis stays true, because $8,9,10 \in S$
Lets assume that $x \in S$. That means $x \in a_{n} \lor x \in b_{n} \lor x \in c_{n}$.
If $x \in a_n$ then $x+1 \in b_x$,
If $x \in b_x$ then $x+1 \in c_x$,
If $x \in c_x$ then $x+1 \in a_x$.
I can't prove that but it's obvious. What do you think about this?
| Write $x=8n+k$ for $0\leq k<8$ and $n\geq 1$. Because $8n=(3+5)n$, the problem is clear if $k=0,3,5$. Let's consider the remaining 5 cases:
\begin{align*}
k=1 & \implies 8n+1=5(n-1)+3(n+2),\\
k=2 & \implies 8n+2=5(n+1)+3(n-1),\\
k=4 & \implies 8n+4=5(n-1)+3(n+3),\\
k=6 & \implies 8n+6=5n+3(n+2),\\
k=7 & \implies 8n+7=5(n+2)+3(n-1).
\end{align*}
Note that $k=4$ is based on $k=1$ (just add one more $3$) and, similarly, $k=6$ and $k=7$ follow from $k=1$ and $k=2$, respectively. So in fact, we have really only considered 2 cases.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $7$ divides $13^n- 6^n$ for any positive integer I need to prove $7|13^n-6^n$ for $n$ being any positive integer.
Using induction I have the following:
Base case:
$n=0$: $13^0-6^0 = 1-1 = 0, 7|0$
so, generally you could say:
$7|13^k-6^k , n = k \ge 1$
so, prove the $(k+1)$ situation:
$13^{(k+1)}-6^{(k+1)}$
$13 \cdot 13^k-6 \cdot 6^k$
And then I'm stuck....where do I go from here?
| There's always this:
$$\begin{split}
x^2-y^2 &= (x-y)(x+y)\\
x^3-y^3 &= (x-y)(x^2+xy+y^2)\\
x^4-y^4 &= (x-y)(x^3 +x^2y+xy^2+y^3)\\
&\vdots\\
x^n-y^n &= (x-y)\left(\sum_{i=1}^{n}x^{n-i}y^{i-1}\right) \mid n\in \mathbb{N} \end{split}$$
In the above, let $x=13$ and $y=6$. Then you can visually see that no matter what $n\in \mathbb{N}$ is, $ 7\mid 13^n-6^n$. And more generally, $x-y \mid x^n-y^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $x^2+4x+18\equiv 0\pmod{49}$ has no solution My method was to just complete the square:
$x^2+4x+18\equiv 0\pmod{49}$
$(x+2)^2\equiv -14\pmod{49}$
$x+2\equiv \sqrt{35}\pmod{49}$
So $x\equiv\sqrt{35}-2\pmod{49}$, which has no real solutions.
I feel that this may be too elementary, is this the correct way to solve this?
| $$(x+2)^2 \equiv -14 \equiv 35 \pmod{49} \text{ is solvable}$$
$$\iff (-14)^{\phi(49)/2} \equiv 35^{\phi(49)/2} \equiv 1 \pmod{49} \text { (But this is false.)}$$
Theorem: If $m$ is has a primitive root (2, 4, and odd prime powers), $x^k \equiv a \pmod{m}$ has a solution if and only if $$\Large a^{\frac{\phi(m)}{gcd(\phi(m),k)}} \equiv 1 \pmod{m}$$ In that case, the equation has $gcd(k, \phi(m))$ incongruent solutions.
Another way: $(x+2)^2 \equiv 35 \pmod{49} \implies (x+2)^2 \equiv 35 \equiv 0 \pmod{7}$
$\implies (x+2)^2 \equiv 0 \pmod{49} \text{ (contradiction)}$
One more way: Use Hensel's lemma to solve for $\text{(some polynomial)} \equiv 0 \pmod{prime^k}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Parity of a Permutation Determine the parity of the following.
$$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 7 & 4 & 1 & 6 & 5 & 3 \end{pmatrix}$$
Workings:
$\sigma = \begin{pmatrix} 1 & 2 & 7 & 3 & 4 \end{pmatrix}\begin{pmatrix} 5 & 6 \end{pmatrix}$
$\sigma = \begin{pmatrix} 1 & 4 \end{pmatrix} \begin{pmatrix} 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 7 \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 5 & 6 \end{pmatrix}$
The parity is odd.
I'm wondering if this is correct.
| If $\tau=\beta_{1}\cdots \beta_{k}$ where each $\beta_{i}$, for $1\le i \le k$, is a transposition, then we have $$sgn(\tau)=(-1)^{k}$$
But, for $\sigma$ we have $k=5$. Hence, $$sgn(\sigma)=(-1)^{5}=-1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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how to show $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ I know that $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ is correct. But having some hard time proofing it using trig relations. Some of the relations I used are
$$
\sin x \cos x= \frac{1}{2} \sin(2 x)\\
\sin^2 x + \cos^2 x= 1\\
\sin^2 x = \frac{1}{2} - \frac{1}{2} \cos(2 x)\\
\cos(2 x) = 2 \cos^2(x)-1
$$
And others. I tried starting with multiplying numerator and denominator with either $\cos x$ or $\sin x$ and try to simplify things, but I seem to be going in circles.
Any hints how to proceed?
| Rearrange and Cross multiply
it is required to prove that
$$\dfrac{\cos x +1 +\sin x}{1-\sin x +\cos x} = \dfrac{1+\sin x}{\cos x}$$ or
$$ \cos^2 x + \cos x (1 +\sin x ) = (1-\sin^2 x) + \cos x (1 +\sin x ) $$ or
$$ \cos x (1 +\sin x ) = \cos x (1 +\sin x ). $$
| {
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Find all possible values of $ a^3 + b^3$ if $a^2+b^2=ab=4$. Find all possible values of $a^3 + b^3$ if $a^2+b^2=ab=4$.
From $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)(4-4)=(a+b)0$. Then we know $a^3+b^3=0$.
If $a=b=0$, it is conflict with $a^2+b^2=ab=4$.
If $a\neq0$ and $b\neq0$, then $a$ and $b$ should be one positive and one negative. This contradict with $ab$=4.
I don't know the solution.
Any help please.
| It is given that $a^{2} + b^{2} = ab =4$ and the goal is to determine all values of $a^{3} + b^{3}$. Now, it is seen that $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2}) = 0$. So far the set of equations becomes
\begin{align}
a^{3} + b^{3} &= 0 \\
a^{2} + b^{2} &= 4 \\
ab &= 4.
\end{align}
From the first, multiply by $a^{3}$ to obtain $0 = a^{6} + (ab)^{3} = a^{6} + 2^{6}$ which yields $a = \pm 2 e^{\pi i/6 \pm 2n \pi i}$. Since $ab=4$ then yields $b = \pm 2 e^{-\pi i/6 \pm 2n\pi i}$. In both cases $n \geq 0$. The possible values of $a$ are
\begin{align}
a \in \{ 2 e^{\pi i/6 + 2n \pi i}, 2 e^{\pi i/6 - 2n \pi i}, - 2 e^{\pi i/6 + 2n \pi i}, - 2 e^{\pi i/6 - 2n \pi i} \}
\end{align}
and for $b$
\begin{align}
b \in \{ 2 e^{-\pi i/6 + 2n \pi i}, 2 e^{-\pi i/6 - 2n \pi i}, - 2 e^{-\pi i/6 + 2n \pi i}, - 2 e^{-\pi i/6 - 2n \pi i} \}
\end{align}
In view of these two sets of values it is seen that the sets must be taken in order of signs, ie the sets $(a, b) \in \{(+,+), (-,-)\}$,
\begin{align}
a^{3} + b^{3} &= 2 (\pm 2)^{3} \cos\left(\frac{\pi}{2}\right) = 0 \\
a^{2} + b^{2} &= 2 (\pm 2)^{2} \cos\left(\frac{\pi}{3}\right) = 4 \\
ab &= (\pm 2)^{2} = 4.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
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